Let $AO$ intersect the circumcircle at $X$. Since $AX$ is a diameter, $ABX$ is a right angle, hence $BX \parallel CH$ (both are perpendicular to $AB$). For the same reason, $CX \parallel BH$, so $BHCX$ is a parallelogram. This means that $XH$ passes through $D$, the midpoint of $BC$. Hence $X$ coincides with $P$ (it lies on both $AO$ and $HD$), and $D$ is also the midpoint of $HX = HP$.
But this implies that $AD$ is a median in both triangles $ABC$ and $AHP$. Since the centroid always divides the median in a $2 : 1$ ratio, we conclude that the centroids of $ABC$ and $AHP$ coincide as well.

It is well known that the reflection $H'$ of $H$ about the side $BC$ lies on the circumcircle. If we reflect $H'$ again about the perpendicular bisector of $BC$, we obtain another point $H''$ on the circumcircle. The double reflection amounts to a single reflection with centre $D$, since $D$ is the intersection of $BC$ and its perpendicular bisector. Hence $H''$ lies on $HD$. Moreover, since $AH'H''$ is a right angle by construction, $AH''$ is a diameter, so $H''$ lies on $AO$ and must thus coincide with $P$. One could now conclude as in the previous solution, but let us show a different approach instead: the centroid of $ABC$ is known to lie on the Euler line, which also passes through $H$ and $O$. On the other hand, $OH$ is also a median in $AHP$ by our observations. $AD$ is a common median in both triangles, hence the intersection of $OH$ and $AD$ is the centroid of both triangles, which completes the proof.