Notice that the condition is equivalent to prove that $A, H, D$ collinear. See that $E=PH\cap TA$ lies on the circle with diameter $AH$, let $F, G$ be the feet of the altitudes from $B, C$ on $\triangle ABC$, so $AEGHF$ is cyclic, and since $TA$ is external bisector we get $\angle TAF=\angle EHF=\angle EAG=\angle EHG$ with this we have $\angle BDC=\angle BHC$ and also that $HP$ and $DP$ are bisectors on $\triangle BHC, \triangle BDC$ resp. Then $\frac{BP}{BC}=\frac{HB}{HC}=\frac{DB}{DC}$, then $\triangle BHC \cong \triangle BDC$, which is $HD\perp BC$ and the result follows.