Prove $|\mathcal P(A)|+|\mathcal P(B)|= |\mathcal P(A)\cup \mathcal P(B)|$ using addition counting principle Let $A$ and $B$ be subsets of the finite set $S$ with $S = A\cup B$ and $A\cap B=\emptyset$. Denote by $\mathcal{P}(X)$ the power set of $X$ and denote by $|Y|$ the number of elements in the set $Y$.
Given a statement $|\mathcal P(A)|+|\mathcal P(B)|= |\mathcal P(A)\cup \mathcal P(B)|$.
Use the Addition Counting Principle to prove or disprove the statement.

I understand that its asking me to find the elements of $\mathcal P(A)$ and $\mathcal P(B)$, but where does $\mathcal{P}(X)$ and $|Y|$ fit in to solve this question?
Well, by the principle of inclusion-exclusion:

$|X\cup Y| = |X|+|Y|-|X\cap Y|$

for finite sets $X,Y$. So in general, your statements does only hold if $P(A)$ and $P(B)$ are disjoint.
I just had this problem myself, and thought up of an answer.

(also posting this for others looking for it)

the $X$ and $Y$ you are questioning are just general terms to describe the functions of modulus (size) $\lvert Y \rvert$, and power set $P(X)$.

The assumption given is that sets $A$ and $B$ are disjoint (hence, their intersection is the empty set), so they have no items in common. Given what we know about the Addition Counting Principle, if 2 sets are finite and disjoint, $\lvert A \cup  B \rvert = \lvert A \rvert + \lvert B \rvert$, and is finite.

In order to prove/disprove $\lvert P(A) \rvert + \lvert P(B) \rvert =  \lvert P(A) \cup  P(B) \rvert$ we must also think about what $P(A)$ and $P(B)$ have in common, just like with sets $A$ and $B$.
We know that the classes $P(A)$ and $P(B)$ are made of sets that are disjoint, so they will have no subsets that are the same. There is only $1$ element shared in both, because that element is also an element of all sets.

So, because $P(A)$ and $P(B)$ share at most $1$ element, they are not disjoint and we cannot conclude that $ \lvert P(A) \rvert + \lvert P(B) \rvert = \lvert P(A) \cup P(B) \rvert$.

Here's an example that I used to convince myself of this.
$S = \{1,2,3,4,5\}$,
$A = \{1,2,3\}$, $\lvert A \rvert = 3$
$B = \{4,5\}$, $\lvert B \rvert = 2$

$\lvert A∪B \rvert = \lvert A \rvert + \lvert B \rvert = 3 + 2 = 5$
so, $A \cup B = \{1,2,3,4,5\} = S$, $\lvert S \rvert = 5$.

$P(A) = \{\emptyset, A, \{1\}, \{2\}, \{3\}, \{1,2\}, \{1,3\}, \{3,2\}\}$,
$\lvert P(A)\rvert = 2^{\lvert A \rvert} = 2^3 = 8$ elements.

$P(B) = \{\emptyset, B, \{4\}, \{5\}\}$,
$\lvert P(B) \rvert = 2^{\lvert B \rvert} = 2^2 = 4$ elements.

$P(A) \cup P(B) = \{∅, A, B, \{1\}, \{2\}, \{3\}, \{1,2\}, \{1,3\}, \{3,2\}, \{4\}, \{5\}\}$,
$\lvert P(A) \cup P(B) \rvert = 11$ elements; as union does not take duplicates, as in $\emptyset$.

so $\lvert P(A) \cup P(B)\rvert \neq \lvert P(A)\rvert + \lvert P(B) \rvert$, as in $8 + 4 = 12 \neq 11$.