The point $P$ has many well-known properties, including the property that $\angle BAP=\angle ACP$ and $\angle CAP=\angle BAP$. We prove this for completeness. Invert at $A$ with radius $\sqrt{AB \cdot AC}$ and reflect about the $A$-angle bisector. Let $P^{\prime}$ be the image of $P$. The angle conditions translate to - $P^{\prime}$ lies on line $AM$ - $P^{\prime}$ lies on the line parallel to $BC$ that passes through the reflection of $A$ about $BC$ (since $P$ lies on the circle with diameter $\overline{AO})$ In other words, $P^{\prime}$ is the reflection of $A$ about $M$. Then $BP^{\prime} \| AC$ and $CP^{\prime} \| AB$, so the circumcircles of $\triangle ABP$ and $\triangle ACP$ are tangent to $AC$ and $AB$, respectively. This gives the desired result. Extend $BP$ and $CP$ to meet the circumcircle of $\triangle ABC$ again at $B^{\prime}$ and $C^{\prime}$, respectively. Then $\angle C^{\prime}BA=\angle ACP=\angle BAP$, so $BC^{\prime} \| AP$. Similarly, $CB^{\prime} \| AP$, so $BCB^{\prime}C^{\prime}$ is an isosceles trapezoid. In particular, this means $B^{\prime}P=CP$, so $BP+PC=BB^{\prime}$. Now observe that $\angle ABP=\angle CAP=\angle BAM$, so if $AM$ meets the circumcircle of $\triangle ABC$ again at $A^{\prime}$, then $AA^{\prime}=BB^{\prime}$. Thus the perimeter of $\triangle BPC$ is $BP+PC+BC=BB^{\prime}+BC=AA^{\prime}+BC$. Now we compute. We have $$BC=2 \sqrt{AO^{2}-OM^{2}}=2 \sqrt{81 \cdot 25}=90$$ and Power of a Point gives $$MA^{\prime}=\frac{BM^{2}}{AM}=\frac{45^{2}}{75}=27$$ Thus $AA^{\prime}+BC=75+27+90=192$.