That's not true

@Above Thanks a lot! I have corrected

Invert around $A$

Redefine $P$ as the $A$-Humpty point. Then $Q \in AD$, as $P$ lies on the $A$-Apollonius circle. Now, it's an easy angle chase to show that $\angle BCQ+\angle CBQ=90^{\circ}$.

Here is my solution for this problem Solution We define again point $P$ is $A$ - Humpty point of $\triangle$ $ABC$; $Q$ is intersection of internal bisector of $\widehat{ABP}$ with $AD$ Since: $P$ is $A$ - Humpty point of $\triangle$ $ABC$, we have: $\widehat{DBP}$ = $\widehat{DAB}$ So: $DB$ tangents ($ABP$) at $B$ or $\dfrac{DP}{DA}$ = $\dfrac{BP^2}{AB^2}$ = $\dfrac{PQ^2}{AQ^2}$ Similarly: $\dfrac{DP}{DA}$ = $\dfrac{CP^2}{AC^2}$ Then: $\dfrac{PQ}{AQ}$ = $\dfrac{CP}{AC}$ or $CQ$ is internal bisector of $\widehat{ACP}$ Hence: $\widehat{BQC}$ = $\widehat{BQD}$ + $\widehat{CQD}$ = $\widehat{BAQ}$ + $\widehat{ABQ}$ + $\widehat{CAQ}$ + $\widehat{ACQ}$ = $\widehat{DBP}$ + $\widehat{PBQ}$ + $\widehat{DCP}$ + $\widehat{PCQ}$ = $\widehat{DBQ}$ + $\widehat{DCQ}$ or $\widehat{BQC}$ = $90^o$

This reminded me a lot of the geometry problem from IMO 1996. Note that the final condition is equivalent with $\frac{AB}{BP}=\frac{AC}{CP}$. Now invert through $A$ with arbritrary radius. We will use " ' " for the images of the points after inversion. Let $\angle ABQ=a$, $\angle ACQ=b$. Note that $\angle B'Q'C'=a+b$. The condition $BQC=90$ will imply $AB'Q'+AC'Q'=270$. So $\angle A=90-a-b$. But $B'P'C'=2a+2b$. As $AP'$ will become under the inversion the symmedian of the triangle $AB'C'$, since the angles imply that $\angle B'P'C'=180-2\angle A$, we'll have that $P'$ is the intersection of the tangents in $B',C'$ wrt $AB'C'$, so $B'P'=C'P'$ Since $B'P'=\frac{r^2 \cdot BP}{AB \cdot AP}=\frac{r^2 \cdot CP}{AC \cdot AP} \implies \frac{AB}{BP}=\frac{AC}{CP}$, the conclusion follows.

Let $R$ be the reflection of $P$ to $D$. Let $CQ$ meet $AD$ at $Q'$. $PCRB$ is a parallelogram. Denote $\angle ABQ=\alpha$ $90=\angle ABQ+\angle CAB+\angle QCA=\alpha+\angle CAB+\angle QCA \implies \angle QCA=90-\alpha-\angle A$ $\angle ABP=2\alpha$ and $\angle PCA=180-2\alpha-2\angle A\implies \angle BRC=\angle CPB=180-\angle A \implies ABRC$ is cyclic. $\angle DCQ=\angle CQD=A+C+\alpha-90=\angle CAR+\angle Q'CA=\angle CQ'D \implies Q'=Q$ as desired.