Power sets of the set difference and difference of the power sets

Question

There are two statements:

$(a)\space\mathcal P(A\setminus B)\subseteq \mathcal P(A)\setminus \mathcal P(B)$

$(b)\space \mathcal P(A\setminus B)\supseteq \mathcal P(A)\setminus \mathcal P(B)$

Generally speaking: $\emptyset\in(\mathcal P(A)\cap \mathcal P(B))$

Is it correct to say power sets can be disjunctive because the empty set is disjunctive to itself? $$\emptyset\cap\emptyset=\emptyset, edited$$ My counter-example for the statement: $(a)\space\mathcal P(A\setminus B)\subseteq \mathcal P(A)\setminus \mathcal P(B)$ is a pair of two disjunctive sets $A$ and $B$ so my claim is equivalent to: $$\mathcal P(A)\nsubseteq\mathcal P(A)\setminus \{\emptyset\}$$ but is kind of a contradiction with the question above.

The counter-example for the statement:$(b)\space \mathcal P(A\setminus B)\supseteq \mathcal P(A)\setminus \mathcal P(B)$ is a pair of sets $A$ and $B$ such that: $$A\cap B\neq \{\emptyset\}$$ /this should be $\emptyset$ instead of $\{\emptyset\}$, as noticed in comments/ I also took into consideration the cardinality of the power sets, but it was insignificant when dealing with disjunctive sets. It works better for a Cartesian product.

Answer 1

Counterexamples:

1. The empty set is in $\mathcal P(A \setminus A)$ while $\mathcal P(A) \setminus P(A)$ is empty.
2. The set $\{0,\pi\}$ is in $\mathcal P(R) \setminus \mathcal P(Q)$ but not in $\mathcal P(R \setminus Q)$.