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.

You play a coin flip game with your friend, but you know that your friend somehow tends to toss heads almost every time. So, you can say something like "Ha I know my friend (prior belief), he always tosses heads, so the probability of him toss heads again will be somewhere between $[0.7,0.9]$", i.e. $\theta$ can be whatever value you want inside that interval.
A natural way for saying that in a probabilistic language, that the probability of success (for your friend) lies inside $[0.7,0.9]$ is to let $\theta$ be a random variable.
Now that $\theta$ is a random variable you can assign it a distribution. But your distribution has to reflect your prior belief, that the probability of success for your friend will be inside the interval $[0.7,0.9]$.
A good choice for distribution for $\theta$ would be a $Beta(a,b)$ distribution (as it takes values inside $[0,1]$ where probabilities also do).
However, this $Beta(a,b)$ distribution must give more attention to values inside $[0.7-0.9]$ which is your prior belief right, that your friend always toss heads.
To do that you can center the distribution around $0.8$ which is the midpoint of the interval $[0.7-0.9]$
You can do that by solving $\frac{a}{a+b}=0.8$ a potential solution for that can be choosing $a=10$ and then $b=2.5$.
So, $\pi(\theta)= Beta(\theta;10,2.5)$ reflect your prior belief about the success probability of your friend that lies inside the interval $[0.7-0.9]$.
Now if you want to say something about as $n$ (the number of samples tends to infinity) then check that the mean of the posterior is
$$Mean = \frac{a+\sum x}{a + \sum x + b +n - \sum x} = \frac{a}{b+n} + \frac{\sum x}{b+n}$$
where for $n\rightarrow \infty$, the mean of your posterior belief $\pi(\theta|x)$ goes to $\bar{x}$
