Just apply Menelaus' theorem on the triangle BMC with the transversal AP You will get OM/PC=1/2

Denote the midpoint $N$ of $[AP]$ . Observe that $MN\parallel BC$ , $OM=MS=\frac 12\cdot PC$ $\implies$ $\frac {OM}{PC}=\frac 12$ .

From the weird conditions it's obvious we need/probably intended to add a point. We reflect $B$ over $M$ to create $D$.Now we get parallelogram $ABCD$. We chase some angles: $\angle CPB = \angle BOP = \angle DOA = \angle OAD$ $\implies$ $AD=OD$ $(1)$. We lenght chase : $\boxed {OM}=BM-BO=BM-BP=\boxed {MD-BP}$ and $\boxed{PC}=BC-BP=AD-BP=OD-BP=(MD+OM)-BP=(MD+(MD-BP))-BP=\boxed{2(MD-BP)}$. We plug this into $\frac{OM}{PC}=\frac{MD-BP}{2(MD-BP)}=\boxed{\frac{1}{2}}$