How do Bayesians interpret $P(X=x|\theta=c)$, and does this pose a challenge when interpreting the posterior? I have seen the post Bayesian vs frequentist interpretations of probability and others like it but this does not address the question I am posing.  These other posts provide interpretations related to prior and posterior probabilities, $\pi(\theta)$ and $\pi(\theta|\boldsymbol{x})$, not $P(X=x|\theta=c)$.  I am not interested in the likelihood as a function of the parameter and the observed data, I am interested in the interpretation of the probability distribution of unrealized data points.
For example, let $X_1,...,X_n\sim Bernoulli(\theta)$ be the result of $n$ coin tosses and $\theta\sim Beta(a,b)$ so that $\pi(\theta|\boldsymbol{x})$ is the pdf of a $Beta(a+\sum x,b + n - \sum x)$.
How do Bayesians interpret $\theta=c$?  $\theta$ of course is treated as an unrealized or unobservable realization of a random variable, but that still does not define or interpret the probability of heads. $\pi(\theta)$ is typically considered as the prior belief of the experimenter regarding $\theta$, but what is $\theta=c$?  That is, how do we interpret a single value in the support of $\pi(\theta)$?  Is it a long-run probability?  Is it a belief?    How does this influence our interpretation of the prior and posterior?
For instance, if $\theta=c$ and equivalently $P(X=1|\theta=c)=c$ is my belief that the coin will land heads, then $\pi(\theta)$ is my belief about my belief, and in some sense so too is the prior predictive distribution $P(X=1)=\int\theta\pi(\theta)d\theta=\frac{a}{a+b}$.  To say "if $\theta=c$ is known" is to say that I know my own beliefs.  To say "if $\theta$ is unknown" is to say I only have a belief about my beliefs.  How do we justify interpreting beliefs about beliefs as applicable to the coin under investigation?
If $\theta=c$ and equivalently $P(X=1|\theta=c)=c$ is the unknown fixed true long-run probability for the coin under investigation: How do we justify blending two interpretations of probability in Bayes theorem as if they are equivalent?  How does Bayes theorem not imply there is only one type of probability?  How are we able to apply posterior probability statements to the unknown fixed true $\theta=c$ under investigation?
The answer must address these specific questions.  While references are much appreciated, the answers to these questions must be provided.  I have provided four Options or proposals in my own solution below as an answer, with the challenges of interpreting $P(X=x|\theta=c)$ as a belief or as a long-run frequency.  Please identify which Option in my answer most closely maps to your answer, and provide suggestions for improving my answer.
I am not writing $P(X=x|\theta=c)$ to be contemptuous.  I am writing it to be explicit since $P(X=x|Y=y)$ is not the same thing as $P(X=x|Y)$.  One might instead be inclined to write in terms of a sample from the prior and use an index of realizations of $\theta$.  However, I do not want to present this in terms of a finite sample from the prior.
More generally, how do Bayesians interpret $P(X=x|\theta=c)$ or $P(X\le x|\theta=c)$ for any probability model and does this interpretation pose any challenges when interpreting $P(\theta=s|\boldsymbol{x})$ or $P(\theta\le s|\boldsymbol{x})$?
I've seen a few other posts tackle questions about Bayesian posterior probability, but the solutions aren't very satisfying and usually only consider a superficial interpretation, e.g. coherent representations of information.
Related threads:
Examples of Bayesian and frequentist approaches giving different results
Bayesian vs frequentist interpretations of probability
UPDATE:
I received several answers.  It appears that a belief interpretation for $P(X=x|\theta=c)$ is the most appropriate under the Bayesian paradigm, with $\theta$ as the limiting proportion of heads (which is not a probability) and $\pi(\theta)$ representing belief about $\theta$.  I have amended Option 1 in my answer to accurately reflect two different belief interpretations for $P(X=x|\theta=c)$.  I have also suggested how Bayes theorem can produce reasonable point and interval estimates for $\theta$ despite these shortcoming regarding interpretation.

Let us think about what you are attempting to ask.  If we define $X=x\in\chi, \chi$ being the sample space, as observed, then in Bayesian thought, it is a constant.  It is an observable.  Instead of using the language of parameters and data, we can think in terms of observables and unobservables.  There is no randomness here.
$\theta$ is normally a random variable in the parameter space, but it is now a constant.  It is crucial that we know how it became so.  It appears from the language of your posting that we are conditioning on it in the likelihood function so that, for our purposes, the likelihood is now $.82^1$.  So it is not a random variable either.  So it is senseless to talk about a probability when everything is a constant.  It would be like discussing the probability that $2+2=4$.
It isn’t impossible to discuss this, but it is difficult for several reasons.  First, the interpretation can change depending on the axioms used to derive Bayes rule.  For example, if we are conditioning on $\theta=.82$, then de Finetti’s axioms would require a prior with mass only on .82.  What if we were using some other axiomatization and the mass of the prior was zero at the point we conditioned it?  Cox’s axioms would find that problematic as well.  Savage’s might not if we allowed for time inconsistency, though why you would change your mind on the prior and not the likelihood is beyond me.
We also need a better definition of what a constant is.  For example, conditioning some parameters on constants is not that unusual in Bayesian thinking.  Sometimes you do know one of them.  There is another case, though, that wrenches up even the Frequentist toolset.
To give an example, the speed of light is known precisely.  However, as distance is now normed against the speed of light, distance is uncertain.  We used to measure the speed of light with uncertainty; we now measure distance with uncertainty.
Let us imagine we get out our carefully built scientific equipment and decide to measure out five kilometers for our morning run.  Our equipment is accurate to within plus or minus twenty meters.  When our device measures five kilometers, we know it is somewhere within 4980 and 5020 meters in reality.  It is close enough.  If this is part of our measuring, we could condition on it being five kilometers as it is close enough for our purposes.  It is also definitely wrong.  Because distance is a value in the real numbers, the probability that our actual distance is five kilometers when it registers five kilometers is a measure zero event.  Our conditioning is wrong with certainty.
A second issue with this type of conditioning problem is a non-mathematical issue.  If, instead, we were running a wrecking ball and hit our intended building, plus or minus twenty meters, we could be hitting the wrong building.  At the same time, we have conditioned our uncertainty away.  Had our wrecking ball been run by a robot, a la E.T. Jaynes, we would have no way to know our decision process was bad.
On the surface, you may think that would not matter, but de Finetti’s coherence wrecks that idea if we are gambling money.  A bad constant could create a Dutch Book.
As I see it, there is no randomness in your problem.  We observed the outcome; it is a certainty.  We observed the parameter.  It is being treated as a certainty.  We are being bigoted in our conditioning in that we are saying there is no uncertainty.
What is probability in the face of perfect certainty?  What do you mean by your question?
