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.
Best Answer
Counterexamples: