By Reim's theorem, $ST\parallel CD$. Note that $\triangle OMN$ and $\triangle PAB$ have two pairs of parallel sides. Thus $P,Q,O$ are colinear if and only if these two triangle are homothetic. Let $M_1, N_1$ be midpoints of $AC, BD$. Clearly $OMSM_1$ is rectangle thus $OM = M_1S$. Similarly, $ON = N_1T$. Now we proceed to prove each part of the problem. If Direction This is the easy one. If $S$ is midpoint of $PC$, then $T$ is midpoint of $PD$. Thus $M_1S = 0.5(AC-PC) = 0.5AP$ and $N_1T = 0.5PB$. Thus $\triangle OMN$ and $\triangle PAB$ are homothetic with ratio $1 : 2$. Only if direction Most solutions are length bashing, but here I will present my clean approach. Animate $S$ along $PC$ with constant velocity. Then $T$ moves linearly along $PD$. Thus the equation $\tfrac{M_1S}{N_1T} = \tfrac{AP}{PB} = \text{constant}$ is a linear equation. That means either there is unique solution or every point on $PC$ is a solution. Suppose that the latter happens, then take $S=C$ and $T=D$. Thus $\tfrac{M_1C}{N_1D} = \tfrac{AC}{BD} = \tfrac{AP}{PB}$ hence $AP : AC = BP : BD$ or $AB\parallel CD$, contradiction. Hence we are done.