It is possible to find additional assumptions on the sets $A$ and $B$ so that in fact $P(A)\cup P(B) = P(A\cup B)$.
Prove that for all sets $A$ and $B$, if $A=B$ then $P(A)\cup P(B) = P(A\cup B)$.
Solution: Suppose $A=B \implies P(A) = P(B) \implies P(A) \cup P(B) = P(A)$ and $P(A \cup B) = P(A \cup A) = P(A)$.
Thus, $P(A) \cup P(B) = P(A) = P(B) = P(A \cup B)$
By containment (or inclusion), let $X \in P(A \cup B) \implies X \subseteq (A \cup B).$
$\implies X \subseteq (A \cup A) = (B \cup B) \implies X \subseteq A=B$.
$\implies X \in P(A) = P(B) \implies X \in P(A) \cup P(B)$.
$\implies P(A \cup B) \subseteq P(A) \cup P(B)$.
As $P(A \cup B) \supseteq P(A) \cup P(B)$. Thus, $P(A) \cup P(B) = P(A \cup B)$.