Simple coordinate bash: Assume WLOG that $AB=AC=12,BC=12$. Set $M=(0,0),A=(0,8),B=(-6,0),C=(6,0),N=(4,0)$. Then, the circumcenter $O$ of $\triangle ABN$ is the intersection of the lines $x=-1$ and $y=\frac{3}{4}x+\frac{7}{4}$, which is $(-1,2.5)$. By the Angle Bisector Theorem, the incenter $I$ of $\triangle ABC$ is $(0,3)$. Then, it is well known that the inradius of a 3-4-5 triangle is $1$, so the inradius of $\triangle ABM$ is $2$ and therefore the incenter $P$ of $\triangle ABM$ is $(-2,2)$. Clearly, $O$ is the midpoint of $IP$. $\blacksquare$