X OBM

Consider all points $A^{\prime }$ such that $P{A}^{\prime }=3$. They lie on a circle with center $P$. The area of $A^{\prime }BC$ is $\frac{BC}{2}$ times the distance of $A^{\prime }$ from $BC$. That distance is maximal for $A^{\prime }P$ perpendicular to $BC$ (because the distance is the distance of $P$ from $BC$ is $P{A}^{\prime }\sin \theta$, where $\theta$ is the angle between $A^{\prime }P$ and $BC$). Hence $AP$ must be perpendicular to $BC$. Similarly $BP$ must be perpendicular to $AC$, so $P$ must be the orthocenter.