What is $X\cap\mathcal P(X)$? Does the powerset of $X$ contain $X$ as a subset, and thus $X\cap \mathcal{P}(X)=X$, or is $X\cap \mathcal{P}(X)=\emptyset$ since $X$ is a member of the 
$\mathcal{P}(X)$, and not a subset? 

It depends.
For "normal" sets, this intersection will be empty, so in general, your second thought is the correct one.
However, if we have something like:
$X = \{1,2,3, \{2,3\}\}$ then intersecting $X$ with its own power set will actually have an element - $\{2,3\}$
The distinction to be made here is subsets vs elements.
