Prove that $\mathcal P(A \cap B)=\mathcal P(A) \cap \mathcal P (B)$ Probably this is an easy problem, but this time I am not understanding these concepts.
The problem starts with:

Let $\mathcal{P}(A)$ denote the set $\{ x ~|~ x \subseteq A \}$

What I think is correct and that I can say from the above expression is that $x$ is a subset of A.
And then we have a set of questions, for example the first one:
Prove that $\mathcal{P}(A \cap B) = \mathcal{P}(A) \cap
\mathcal{P}(B)$ by showing that the statements $x \in \mathcal{P}(A
\cap B)$ and $x \in \mathcal{P}(A) \cap \mathcal{P} (B)$ are
equivalent.
The only thing that comes in mind is to replace the expression $\mathcal{P}(A \cap B) $ with its equivalent of this $\{ x ~|~ x \subseteq A \}$, but I am not really understanding the process and the logic...
 A: You are going in the right direction. $\mathcal{P}(A) = \{x | x \subseteq A \}$, so just by replacing the symbols we see $\mathcal{P}(A \cap B) = \{x | x \subseteq A \cap B\}$.  So you are being asked to relate
$$\{x | x \subseteq A \cap B\}$$
To $\mathcal{P}(A) \cap \mathcal{P}(B)$ which is
$$\{x | x \in \mathcal{P}(A) \text{ and } x \in \mathcal{P}(B)\}$$
Does this help you get going?
edit: I want to comment that normally to show two sets are equal the foolproof approach is to show that they are subsets of each other.  Playing with the "rules" of a set to show it is equivalent to another should give you all the information you  need to do this, but if you want to dot your i's for this you ought to show each subset relationship.
A: hint: to get from 
$$
\{x|x⊆A∩B\}
$$
to 
$$
\{x|x∈P(A) \text{ and }x∈P(B)\}
$$
The property $$
X\subset Y \iff X\cup Y = Y
$$can be useful.

solution:
$$\begin{align}
\{x|x∈P(A) \text{ and }x∈P(B)\}
&= \{x|x\subset A \text{ and }x\subset B\}\\
&= \{x|x\cup A = A \text{ and }x\cup B = B\}\\
&= \{x|x\cup A \cup B = A\cup B\}\\
&= \{x|x⊆A∩B\} 
\end{align}$$
