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