Solution:
No, it is not true that for any sets $A$ and $B, P(A) \cup P(B)=P(A \cup B)$
Justification:
Let’s suppose,
$A=\{0,1\}$
And, $B=\{1,2\}$
$\therefore \text{ }A~\cup ~B~=\text{ }\left\{ 0,\text{ }1,\text{ }2 \right\}$
As per the question, we have,
$\begin{array}{l}
P(A)=\{\phi,\{0\},\{1\},\{0,1\}\} \\
P(B)=\{\phi,\{1\},\{2\},\{1,2\}\} \\
\therefore P(A \cup B)=\{\phi,\{0\},\{1\},
\end{array}$
$\mathrm{P}(\mathrm{A}) \cup \mathrm{P}(\mathrm{B})=\{\phi,\{0\},\{1\},\{2\},\{0,1\},\{1,2\}\}$
$\therefore P(A) \cup P(B \neq P(A \cup B))$
As a result, the given statement is false.