jmc

Let the circle intersect $\overline{PM}$ at $B$. Then note $△OPB$ and $△MPA$ are similar. Also note that $AM=BM$ by power of a point. Using the fact that the ratio of corresponding sides in similar triangles is equal to the ratio of their perimeters, we have$$\frac{19}{AM}=\frac{152-2AM-19+19}{152}=\frac{152-2AM}{152}$$Solving, $AM=38$. So the ratio of the side lengths of the triangles is 2. Therefore,$$\frac{PB+38}{OP}=2\text{ and }\frac{OP+19}{PB}=2$$so $2OP=PB+38$ and $2PB=OP+19.$ Substituting for $PB$, we see that $4OP-76=OP+19$, so $OP=\frac{95}{3}$ and the answer is $\boxed{98}$.