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?
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 $Y \cap \wp(Y) = \{\{0\}\} \neq \varnothing.$
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' &amp; p(X)

So, X intersection/and P(X)={}