Note. Another problem, similar but different, in the same contest used the same diagram.

It suffices to show $P$ lies inside the circumcircle of $\triangle BHC$. Let $ACXH$ and $ABYH$ be parallelograms. Then $\angle HXC=\angle HAC=\angle CBH$, so $BHCX$ is cyclic and similarly $BHCXY$ is cyclic. So $P$ which lies on $XY$ lies inside the circumcircle.

Let $P'=BC\cap (BHP)$, we would like to show that $P'D<DC$. Note that $$\angle P'PD=\angle P'PH-\angle DPH=\angle P'BH-\angle DPH=\angle HAC-\angle HAD=\angle DAC.$$ Also, $P'D=\tan{\angle P'PD}\cdot PD=\tan{\angle P'PD}\cdot AH$ and by law of sines $$\frac{AD}{\sin{\angle C}}=\frac{CD}{\sin{\angle P'PD}}.$$ Therefore, $$\frac{P'D}{DC}=\frac{AH\cdot \sin{\angle C}}{AD\cdot \cos{\angle DAC}}=\frac{AL}{AR},$$ where $L$ is the foot from $B$ to $AC$ and $R$ is the foot from $D$ to $AC$. As $D$ lies between $B$ and $C$, we conclude that $AL<AR<AC$, therefore indeed $P'D<DC$, we are done.