Let $l_a$ be the bisectrix of $\angle A$. Then $l_a \perp AT$ and $PH \parallel l_a$ (as, by condition, $PH \perp AT$). Further,
$$\angle (BH, HP) = \angle (BH, AC) + \angle (AC, HP) = 90^\circ + \angle (AC, l_a) = 90^\circ + \angle (l_a, AB) = 90^\circ + \angle (HP, AB) = \angle (HP, AB) + \angle (AB, CH) = \angle (HP, CH).$$
Hence, $HP$ is the bisectrix of $\angle BHC$.

Let $AH$ meet the circumcircle of $\triangle ABC$ at $D_1 \neq A$. We have $\angle D_1BC = 90^\circ - \angle C = \angle HBC$ and, similarly, $\angle D_1CB = \angle HCB$. Then the triangles $BHC$ and $BD_1C$ are equal, Hence, $D_1P$ is the bisectrix of $\angle BD_1C$. Therefore, $D_1P$ intersects the circumcircle at the middle of the arc $BAC$, i.e. at the point $T$. So, $D_1 = D$ and $AD \perp BC$. Thus the statement follows from the well-known

Lemma. The diagonals of the quadrilateral $ABCD$ are perpendicular if and only if $AB^2 + CD^2 = AC^2 + BD^2$.