2022/23 AHSMC Part 1 Problem 6

We have $PA + PC \ge AC$, $PB + PD \ge BD$ and the equality in both inequalities is obtained when $P$ is on $AC$ and $BD$ i.e., $P$ is the intersection of the diagonals of the square. Therefore, $$2\sqrt{2} = PA + PC + PB + PD \ge AC + BD = 2\sqrt{2}.$$ Hence $PA + PC + PB + PD = AC + BD$, thus $P$ is the center of the square. The answer is (B).