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? 
 A: Actually: $$A \subset X \iff A \in \wp(X).$$ All situations can happen. If $X = \{  0,1\}$, then $$\wp(X) = \{  \varnothing, \{0\},\{1\},\{0,1\} \},$$ so $X \cap \wp(X) = \varnothing$. But if you take, say $Y = \{ 0, \{0\}  \}$, then: $$\wp(Y) = \{ \varnothing, \{0\}, \{\{0\}\}, \{ 0,\{0\} \}  \},$$ so $Y \cap \wp(Y) = \{\{0\}\} \neq \varnothing.$
A: This can be many things.
If $\varnothing\in X$, then $X\cap\mathcal P(X)$ is definitely not empty, since $\varnothing$ is an element of both sets. And if $X=\varnothing$, then $X\cap\mathcal P(X)=X=\varnothing$.
Moreover, we say that $X$ is a transitive set if whenever $A\in X$, then $A\subseteq X$. Recall, before pondering this definition, that in modern set theory, everything is a set. So this makes sense. If $X$ is a transitive set, then $X\subseteq\mathcal P(X)$, in which case $\mathcal P(X)\cap X=X$.
But, thinking about this naively, or perhaps from a "naive type theoretic" point of view, where the objects inside of $X$ are of type I, and the objects in $\mathcal P(X)$ are sets of objects of type I, and sets of objects of type I, are not objects of type I themselves. For example, sets of natural numbers "are not" natural numbers (type theoretically, sure; but set theoretically this is certainly not necessarily the case). In this naive thinking, indeed $X\cap\mathcal P(X)=\varnothing$ since each holds objects of different a type.
But, again, this depends on how you view your foundation of mathematics, and what exactly is the set $X$. In some ways you can arrange that $X\cap\mathcal P(X)=X$, or at least is non-empty; and in other ways you can arrange that it is not the case.
A: The powerset of $X$ intersects $X$ only when some member of $X$ is also a subset of $X$.
If $X=\{a,b,c\}$ then its power set is $\{\{\}, \{a\}, \{b\}, \{c\},\{a,b\},\{a,c\},\{b,c\},\{a,b,c\}\}$.
If $X=\{a,b,c,\{a,b\}\}$ then $\{a,b\}$ is both a member and a subset of $X$ and thus $X$ and the powerset of $X$ have one member in common --- they intersect.
A: 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.
A: Define a set X={a,b,c} having 3 elements(let).
So, power set of X={{a},{b},{c},{a,b},{a,c},{b,c},{a,b,c},{}}, [i.e it has 2^(3)=8 elements.]
So, X AND P(X) means the common part in 'X' & p(X)
So, X intersection/and P(X)={}
