Well, on a practical and obvious level . $A \times B$ = a set of ordered pairs so $P(A\times B)$ = a set of sets of ordered pairs. $P(A)$ = a set of set of elements. So $P(A) \times P(B)$ = a set of ordered pairs of sets.
It might seem abstract but a set of ordered pairs of sets = {({..},{....})}, is completely different than a a set of sets of ordered pairs. {{(x,y)}}.
e.g.
P({1,2} X {3,4}) = P({(1,3),(1,4),(2,3),(2,4)}) = {$\emptyset$, {(1,3)},{(1,4)},{(2,3)},{(2,4)},{(1,3),(1,4)},{(1,3),(2,3)},{(1,3),(2,4)},{(1,4),(2,3)}{(1,4),(2,4)},{(2,3),(2,4)},{(1,3),(1,4),(2,3)},{(1,3),(1,4),(2,4)},{(1,3),(2,3),(2,4)},{(1,4),(2,3),(2,4)},{(1,3),(1,4),(2,3),(2,4)}}
wherease P({1,2})X P({3,4}) = {$\emptyset$, {1},{2},{1,2})X ($\emptyset$, {3},{4},{3,4}) =
= {($\emptyset, \emptyset$),($\emptyset$, {3}), ($\emptyset$, {4}),($\emptyset$, {3,4}),({1}, $\emptyset$),({1}, {3}), ({1}, {4}),({1}, {3,4}),({2}, $\emptyset$),({2}, {3}), ({2}, {4}),({2}, {3,4})({1,2}, $\emptyset$),({1,2}, {3}), ({1,2}, {4}),({1,2}, {3,4})}
Different things. By coincidence they both have 16 elements.
In general:
$|A \times B| = |A| * |B|$
$|P(A)| = 2^{|A|}$.
So $|P(A \times B)| = 2^{|A||B|}$ while $|(P(A) \times P(B)| = 2^{|A|}2^{|B|}=2^{|A| + |B|} \ne 2^{|A||B|}$
so we know this can't be true.
Best Answer
Your first instinct was right: $P(\{a,b\}) \times P(\{p,q\})$ contains $(\emptyset, \emptyset)$, but not $\emptyset$. $(\emptyset, \emptyset)$ is not the empty set, so $P(a,b) \times P(p,q)$ does not contain the empty set.
And yes, every powerset contains the empty set, but $P(a,b) \times P(p,q)$ is not a powerset, it's the cartesian product of two powersets.