Solution 1 If $PH\perp AB$ ($H\in AB$), we desire $S_{DHP} = S_{BHP}$. Since $\frac{S_{DHP}}{S_{AHP}} = \frac{PD}{PA}$ and $\frac{S_{AHP}}{S_{BHP}} = \frac{AH}{BH}$, we now want $PD \cdot AH = PA \cdot BH$. Since $\triangle AHP \sim \triangle PHQ$ we get $\frac{AH}{PA} = \frac{PH}{QP}$ and so we reduce to $\frac{PH}{BH}=\frac{QP}{PD}$. The latter follows from $\triangle PHB \sim \triangle PQD$, via $\angle PDQ = \angle ABC$ and $\angle DPQ = \angle BHP = 90^{\circ}$.
Solution 2 Let $M$ be the midpoint of $BD$ - we show $PM \perp AB$. This is equivalent to $PA^2 - PB^2 = AM^2 - BM^2$. The median formula in triangle $ABD$ gives $AM^2 = \frac{1}{4}(2AB^2 + 2AD^2 - BD^2)$ and since $BM = \frac{BD}{2}$, we get $AM^2 - BM^2 = \frac{1}{2}(AB^2 + AD^2 - BD^2)$. The cosine law in $ABD$ gives $$AM^2 - BM^2 = AB \cdot AD \cdot \cos \angle BAD.$$and the same law in $APB$ gives $PA^2 - PB^2 = 2 \cdot AP \cdot AB \cdot \cos \angle BAD - AB^2$. So it remains to show
$AB \cdot AD \cdot \cos \angle BAD = 2 \cdot AP \cdot AB \cdot \cos \angle BAD - AB^2$, i.e. $AB = (2AP - AD)\cos\angle BAD$, i.e. $AB = (AP + AD)\cos\angle BAD$. If $PH \perp AB$ ($H\in AB$), we reduce to $AB = AH + PD\cos \angle BAD$, i.e. $BH = PD \cot \frac{AH}{AP}$, i.e. $PD \cdot AH = PA \cdot BH$. Now finish as in Solution 1.