, 2019

We observe that: $P_A \cdot P_B = 8! = c$. By symmetry, without loss of generality, we suppose that $P_A \leq P_B$, and hence $P_A \leq \sqrt{c}$. We write
$$P_A + P_B = P_A + \frac{c}{P_A}$$
and $P_A = x$. We consider the function $f(x) = x + \frac{c}{x}$, with $1 \leq x \leq \sqrt{c}$. The function $f(x)$ is strictly decreasing, because for $1 \leq x < y \leq \sqrt{c}$ it follows that $f(x) > f(y)$. Indeed, for $1 \leq x < y \leq \sqrt{c}$, we have $x - y < 0$, $xy - c < 0$ and
$$f(x) - f(y) = x - y + \frac{c(y-x)}{xy} = \frac{(x-y)(xy-c)}{xy} > 0. \quad (*)$$
Since $x$ is integer and it is not possible to be equal to $\sqrt{c}$, the least possible value is the closest integer to $\sqrt{c}$. We have $\lfloor\sqrt{8!}\rfloor = \lfloor24\sqrt{70}\rfloor = 200$, and so the closest integer is $24 \cdot 8 = 192$. Therefore the least possible value is
$$f(192) = 192 + 210 = 402$$
and it can be approached, for example for the sets $A = \{4,6,8\}$, $B = \{1,2,3,5,7\}$.