China Mathematical Competition

As shown in the figure, assuming that $AC$ and $BD$ meet at point $F$, we have $$|PA|+|PC|\ge |AC|=|FA|+|FC|$$ and $$|PB|+|PD|\ge |BD|=|FB|+|FD|.$$ When $P$ coincides with $F$, $|PA|+|PB|+|PC|+|PD|$ reaches the minimum. That is, $|AC|+|BD|=3\sqrt{2}+2\sqrt{5}$. So $3\sqrt{2}+2\sqrt{5}$ is the required answer.