Since the sides $AB$ and $CD$ of a square are parallel, the cross-lying angles are equal: $\angle PKM_1 = \angle PK_1M$ and $\angle PM_1K = \angle PMK_1$. Therefore the triangles $PKM_1$ and $PK_1M$ are similar and have equal ratios $KP : K_1P = M_1P : MP$. Denote $KP : K_1P = k$, then $MP : PM_1 = 1/k$. Similarly, for $LP : PL_1 = l$ we obtain $NP : PN_1 = 1/l$. Hence the equality given in the problem is equivalent to $k + 1/k + l + 1/l = 4$, which can be transformed as
$$0 = k + \frac{1}{k} - 2 + l + \frac{1}{l} - 2 = \frac{k^2 + 1 - 2k}{k} + \frac{l^2 + 1 - 2l}{l} = \frac{(k-1)^2}{k} + \frac{(l-1)^2}{l}.$$
Clearly, the equality holds if and only if $k = l = 1$ which means that
$$KP = K_1P, \quad LP = L_1P, \quad MP = M_1P \quad \text{and} \quad NP = N_1P$$
(in particular $P$ is the center of the square $ABCD$). Now the required equality is obvious.