Is this a known quantity or a random variable? For two random variables $X_1$, $X_2$, probability of superiority ($PS$) is defined as the probability that a randomly chosen $X_1$ is greater than a randomly chosen $X_2$:
$PS \equiv Pr(X_1 > X_2) \tag{1}$
Suppose $X_1$, $X_2$ follow independent normal distributions with known parameters.
Since the parameters are known,
$PS = \Phi\left(\frac{\mu_1 - \mu_2}{\sqrt{\sigma_1^2 + \sigma_2^2}}\right) \tag{2}$
Specifically, because distribution parameters are known, $PS$ is a single known quantity, not a random variable.
Suppose now that the distribution parameters are not known, but we observe some data on $X_1$, $X_2$. My objective is to find $PS$ conditional on the data. How do we calculate $PS$? I can see two approaches.
Approach 1. The distribution parameters conditional on the data have known sampling distributions (normal for $\mu$ and chi-square for $\sigma^2$). Draw parameters from these known sampling distributions. For each set of parameters drawn, calculate $PS$ per $(2)$.
This gives us a bootstrapped distribution of $PS$. $PS$ is a random variable with a distribution.
Approach 2. Per the definition $(1)$, we are actually interested in the distributions of future data, not in the sampling distributions of the parameters. Conditional on the data (not the parameters), $X$'s have a known distribution. The posterior predictive of each $X$ is the 3-parameter Student t distribution (see here, equation 100).
We could work out the difference between two random variables that have 3-parameter Student t distributions, and the result will be similar to $(2)$ but with heavier tails and, instead of having parameters in the equation, it will have statistics.
Under this approach, because exact distributions of $X_1$, $X_2$ conditional on the data are known, $PS$ turns out to be a single known quantity, not a random variable, just like when distribution parameters were known.
Question. Which approach is correct?
It feels to me like Approach 2 is correct, as it integrates out distribution parameters, which are not even part of the definition of $PS$. The "problem" with this approach is that since $PS$ is known, it has no confidence interval and one cannot do hypothesis tests on it.


$PS \equiv Pr(X_1 > X_2) \tag{1}$

is a statement about two random variables. You describe approach 2 by saying

[...] Conditional on the data (not the parameters), $X$'s have a known distribution. [...]

This is incorrect. You seem to be describing the Bayesian approach to estimating the parameters of distributions of random variables and the priors. It is not known but estimated unless you mean something else than you described.
Then you comment

In the second approach, $PS$ appears to be a known quantity. It's not a random variable, because it is conditional on the data, which is not random. It doesn't have a distribution. It's just a number.

In statistics, we consider the samples as realizations of random variables. If you are concerned about "just numbers" then statements like $PS \equiv Pr(X_1 > X_2)$ make no sense. You gathered some data that presumably is a sample, so it is random due to the sampling. Most likely you want to generalize what you learned from the data to some population, so again, if you treat the data as "just numbers" that tell you about the "known" parameters, then you cannot generalize the results to the population without considering them as random variables.
So answering the question answered in the title, "known quality" is something that you know about the population of interest. If you calculate something from the sample, it is estimated, not known. It would be known only if you have the data about the whole population.
