The circle touches $\overline{AM}$ at $A$ itself, so the tangents from $M$ give $MT = MA,$ where $T$ is the tangency point on $\overline{PM}$

Right triangles $POT$ and $PMA$ are similar, and the small one’s perimeter is $PA + PT = 152 - 2\,MA;$ comparing perimeter ratios gives an equation in $MA$ alone

Let $T$ be the point where the circle touches $\overline{PM}.$ Since $\overline{AM} \perp \overline{AP}$ and $O$ lies on $\overline{AP}$ at a distance from line $AM$ equal to $19,$ the circle is tangent to $\overline{AM}$ at $A$ itself, so the two tangents from $M$ give $MT = MA.$ Right triangles $POT$ and $PMA$ (right angles at $T$ and $A$) share angle $P,$ so they are similar with ratio $\frac{OT}{MA} = \frac{19}{MA}.$

The small triangle’s perimeter is $PO + OT + TP$ $= (PA - 19) + 19 + TP$ $= PA + PT,$ and since $MT = MA,$ $$\begin{aligned} PA + PT &= PA + PM - MT \\ &= 152 - 2\,MA. \end{aligned}$$ Perimeters of similar triangles are in the ratio of similarity, so $\frac{19}{MA} = \frac{152 - 2\,MA}{152},$ which simplifies to $MA^2 - 76\,MA + 1444$ $= (MA - 38)^2$ $= 0.$ Thus $MA = 38.$

The ratio of similarity is then $\frac{19}{38} = \frac{1}{2},$ so $PO = \frac{1}{2} PM.$ From the perimeter, $PA + PM = 152 - 38 = 114,$ and $PA = PO + 19,$ so $\frac{1}{2} PM + 19 + PM = 114,$ giving $PM = \frac{190}{3}$ and $OP = \frac{95}{3}.$ Hence $m + n = 95 + 3 = 98.$