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.

I think that the issue here is whether the likelihood that is used to turn a Bayesian prior into a Bayesian  posterior has a Bayesian interpretation, or whether it is necessarily a long-run frequency.  IF that is the question, then I would say "yes", of course it does.
If we have a Bernoulli likelihood then $f(k;\theta) = \theta^k(1-\theta)^{1-k}$ for $k \in \{0,1\}$, where $\theta$ is a parameter, with unknown "true" probability $c$.  Now I would regard $c$ as a "physical probability" (IIRC Good uses that term as well).  If we are talking about macroscopic objects, like coins, then there is no such thing as "randomness".  Whether a coin comes down heads of tails is entirely deteministic.  The only reason we can't know whether it comes down heads or tails is that we don't have perfect knowledge of the initial condtions.  So a "physical probability" is basically summarising appearance of randomness that are caused by that lack of knowledge.
It is important to distinguish physical probabilities from either Bayesian probabilities or frequentist ones.  It distinct from frequentist probabilities because the long run frequency normally stems directly from the physical probability being (assumed to be) the same for all possible trials.  However, they are not directly equivalent.  The physical probability is directly the probability that a particular coin flip will come up heads, because it describes the physics that makes that so.  A frequentist probability, as a long-run frequency can't be applied to a particular coin flip (as it has no long run frequency, it happens only once). It can only make a probabilistic statement about a (fictitious) population of coin flips, of which this can be considered a particular sample.
Note that something that can only ever happen at most once can have a physical probability of happening, but it can't have a long run frequency.
A physical probability isn't a Bayesian probability either.  Bayesian probabilities represent our beliefs (subjective or objective) about the plausibility of different values of that physical probability.  The Bayesian is making a probabilistic model of the physics, it isn't the physics itself.
So for me, I would view $f(k;c)$ as neither Bayesian nor frequentist, or perhaps both, but a statement about the physics of the date generating process.  It is equally true for single observations, or when looking at a long run of observations.
So in this case, to get our posterior, we would say
$$p(\theta|X=1) = \frac{P(x=1;\theta)p(\theta)}{P(X=1)}$$
Note that $c$ does not appear anywhere in our inference to obtain our posterior belief, which is why I think the question as posed is essentially meaningless.
Of course we could look at our posterior to evaluate $p(c|X=1)$, but this would just give us the point probability of the model parameter, $\theta$, being the same numeric value as the true "physical" probability.  As this is evaluating a specific point on a continuous distribution, it is just a density, not an actual probability.  Probably not very useful.
At the end of the day, it is a very badly posed question, but that seems to be the most meaningful answer I can give to the most informative intepretation that is reasonably consistent with the question as posed.
