Is $\mathcal P(A) \times \mathcal P(B)=\mathcal P(A\times B)$? Let $A=\lbrace 1,2 \rbrace$ and $B=\lbrace 2,3,4 \rbrace$.
Is  $\mathcal P(A)\times \mathcal P(B)=\mathcal P(A\times B)$?
My attempt and reasoning, from the first one, I compute the powerset of both and get $\mathcal P(A)=\bigg\lbrace \lbrace 1 \rbrace, \lbrace 2 \rbrace, \lbrace 1,2 \rbrace,\emptyset \bigg\rbrace$ and $\mathcal P(B)=\bigg\lbrace \lbrace 2 \rbrace, \lbrace 3 \rbrace, \lbrace 4 \rbrace, \lbrace 2,3 \rbrace,\lbrace 2,4 \rbrace, \lbrace 3,4 \rbrace, \lbrace 2,3,4 \rbrace, \emptyset\bigg\rbrace$ and  $|\mathcal P(A) \times \mathcal P(B)|=32 \quad .$
The right hand side I compute $A \times B$ to get $\lbrace (1,2),(1,3),(1,4),(2,2),(2,3),(2,4),\emptyset \rbrace$ It is easy to see that $|\mathcal P(A \times B)|=36$ so they shouldn't be equal by this reasoning right?,
 A: Your counting argument is a good one.  $$|P(A)\times P(B)|=|P(A)||P(B)|=2^{|A|}2^{|B|}=2^{|A|+|B|}$$
while $$|P(A\times B)|=2^{|A||B|}$$
Hence so long as $|A|+|B|\neq |A||B|$, the two sets will be of different sizes.
A: While there are many arguments that $\def\P{\mathcal P}\P(A) \times \P(B)=\P(A\times B)$ cannot be true, I should like to note that it is not even a meaningful mathematical statement in the common mathematical sense, in the same way as $\Bbb R\cap(\Bbb N\times\Bbb Q^3)=\emptyset$ is not a meaningful statement. It is not meaningful because we are comparing things that are not of the same kind (individual real numbers against ordered pairs of a natural number and a $3$-tuple of rationals in the latter example). Concretely, $\P(A) \times \P(B)$ is a set of ordered pairs of sets, while $\P(A\times B)$ is a set of sets of ordered pairs; they don't contain the same kind of elements, so comparing them is not meaningful.
To give meaning to such statements, one would be forced to adopt the strict set-theoretic point of view that everything mathematical is a set (unless it is too large to be so), and to know about the precise set-theoretic constructions used to introduce them (different types of numbers, ordered pairs, maps). In the case of the question one would be forced to interpret ordered pairs as sets. There do exist constructions that define ordered pairs to be certain kinds of sets, but it can certainly be done in more than one way, and in general the answers to this kind of question would depend on the construction chosen (on implementation details, as they would say in Computer Science). See the answer by bof for what one gets from the question using a particular (popular) convention.
A: $P(A\times B)\not\subseteq P(A)\times P(B)$ because $\emptyset\in P(A\times B)$ while $\emptyset\notin P(A)\times P(B)$, at least if you're using Kuratowski's definition of ordered pairs: $(x,y)=\{\{x\},\{x,y\}\}\ne\emptyset$.
A: Showing that $P(A)\times P(B)$ and $P(A \times B)$ have different cardinalities is a good approach, but you did some of the computations incorrectly. 
Since $|P(A)| = 2^{|A|} = 2^2 = 4$ and $|P(B)| = 2^{|B|} = 2^3 = 8$, we have $|P(A)\times P(B)| = |P(A)| \cdot |P(B)| = 4 \cdot 8 = 32$.
Since $|A \times B| = |A| \cdot |B| = 2 \cdot 3 = 6$, we have $|P(A \times B)| = 2^{|A \times B|} = 2^6 = 64$. 
Another way to see that $P(A)\times P(B) \neq P(A \times B)$ is to note that $P(A) \times P(B)$ contains ordered pairs of sets like $(\{2\},\emptyset)$ while $P(A \times B)$ contains sets of ordered pairs like $\{(1,2),(2,3)\}$. Since $P(A)\times P(B)$ and $P(A \times B)$ don't contain the same kinds of objects, they aren't the same (even if they happened to contain the same number of elements). 
A: Your reasoning is correct except for the last step, where the cardinality of $A \times B$ should equal 6 rather than 7 (since the empty set $\phi$ does not belong to $A \times B$).  Therefore the power set of $A \times B$ has cardinality $2^6=64$.  Whereas $\mathcal{P}(A) \times \mathcal{P}(B)$ has cardianlity $2^2 \times 2^3 = 32$.  Hence the two sets are not equal in general. 
