The notation is very misleading with $p(\cdot)$ being heavily overloaded with multiple meanings and complete disregard for the difference between random variables and the values that they take on.
There is a parameter $\theta$ of unknown value which is modeled as a random variable $\Theta$, discrete or continuous depending on what $\theta$ is believed to be. Assuming that $\theta$ is believed to take on discrete values, $$\Theta \sim p_\Theta(\theta)\tag{1}$$ is saying that $\Theta$ is being modeled as a discrete random variable with probability mass function $p_\Theta(\theta)$, or to paint refined gold and gild the lily, the function $p_\Theta(\cdot)$ has the property that $$P\{\Theta = \theta_i\} = p_\Theta(\theta_i).$$ There are $n$ observations $x_1, x_2, \ldots, x_n$ which are also modeled as random variables $X_1$, $X_2$, $\ldots$, $X_n$. Assuming that everything is discrete, on a trial of the experiment, these random variables $\Theta, X_1, X_2, \ldots, X_n$ have values $\theta, x_1, x_2, \ldots, x_n$ of which we can observe only $x_1, x_2, \ldots, x_n$. Now, the conditional probability mass function of the $X_i$ given that $\Theta$ has value $\theta$ is $$p_{X_1, X_2, \ldots, Xn \mid \Theta = \theta}(x_1, x_2, \ldots, x_n \mid \Theta = \theta)\tag{2}$$
meaning that
$$P\big\{X_1 = x_1, X_2 = x_2, \ldots, X_n = x_n \mid \Theta = \theta_i
\big\} = p_{X_1, X_2, \ldots, Xn \mid \Theta = \theta_i}(x_1, x_2, \ldots, x_n \mid \Theta = \theta_i).$$
Notice that $p$ has different subscripts in $(1)$ and $(2)$ to emphasize that there are two different functions being used here; one with just one argument and another with $n$ arguments, a far better arrangement than $p$ meaning different things depending on where it occurs and also better than the "canonical"
Let $X \sim f(x)$ and $Y \sim f(y)$ be random variables $\ldots$
or the physicists'
$X \sim f(X), Y \sim f(Y), X,Y \sim f(X,Y)$
where the density of $X$ can be quite different from the density of $Y$ and one is supposed to figure out from the letter used as the argument of $f\cdot)$ as to whether the density function of $x$ is meant or the density function of $Y$.