Solution:
For the maximum, we use the fact that $(P_{A} - 1)(P_{B} - 1) \geqslant 0$, to get that $P_{A} + P_{B} \leqslant P_{A} P_{B} + 1 = 11! + 1$. Equality holds if and only if $A = \{1\}$ or $B = \{1\}$.

For the minimum observe, first that $P_{A} \cdot P_{B} = 11! = c$. Without loss of generality let $P_{A} \leqslant P_{B}$. In this case $P_{A} \leqslant \sqrt{c}$. We write $P_{A} + P_{B} = P_{A} + \frac{c}{P_{A}}$ and consider the function $f(x) = x + \frac{c}{x}$ for $x \leqslant \sqrt{c}$. Since
$$f(x) - f(y) = x - y + \frac{c(y - x)}{y x} = \frac{(x - y)(x y - c)}{x y}$$
then $f$ is decreasing for $x \in (0, c]$.
Since $x$ is an integer and cannot be equal with $\sqrt{c}$, the minimum is attained to the closest integer to $\sqrt{c}$. We have $\lfloor \sqrt{11!} \rfloor = \left\lfloor \sqrt{2^{8} \cdot 3^{4} \cdot 5^{2} \cdot 7 \cdot 11} \right\rfloor = \lfloor 720 \sqrt{77} \rfloor = 6317$ and the closest integer which can be a product of elements of $X$ is $6300 = 2 \cdot 5 \cdot 7 \cdot 9 \cdot 10$.
Therefore the minimum is $f(6300) = 6300 + 6336 = 12636$ and it is achieved for example for $A = \{2,5,7,9,10\}$, $B = \{1,3,4,6,8,11\}$.

Suppose now that there are different sets $A$ and $B$ such that $P_{A} + P_{B} = 402$. Then the pairs of numbers $(6300, 6336)$ and $(P_{A}, P_{B})$ have the same sum and the same product, thus the equality case is unique for the numbers $6300$ and $6336$. It remains to find all possible subsets $A$ with product $6300 = 2^{2} \cdot 3^{2} \cdot 5^{2} \cdot 7$. It is immediate that $5,7,10 \in A$ and from here it is easy to see that all possibilities are $A = \{2,5,7,9,10\}, \{1,2,5,7,9,10\}, \{3,5,6,7,10\}$ and $\{1,3,5,6,7,10\}$.

Solution:
We have $P_{A} + P_{B} \geqslant 2 \sqrt{P_{A} P_{B}} = 2 \sqrt{11!} = 1440 \sqrt{77}$. Since $P_{A} + P_{B}$ is an integer, we have $P_{A} + P_{B} \geqslant \lceil 1440 \sqrt{77} \rceil = 12636$. One can then follow the approach of the first solution to find all equality cases.