If $AB=AC$, then the problem is clearly true. Assume otherwise. Let $K$ be the second intersection of $(AID)$ with line $BC$ (otherwise known as the foot of the $A$-external bisector). This is our key point: Claim: $\triangle KBI\sim ABP$ and $\triangle KCI\sim \triangle ACQ$. Proof: Just angle-chasing. Note that $\angle ABP=\angle KBI=\angle ABI$ (using directed angles), and $\angle PAB=\angle DAI=\angle DKI=\angle IKB$ by using the fact that the original information implies $\angle PAD=\angle BAI$. Now, note that $\frac{AP}{KI}=\frac{AB}{KB}$ and $\frac{AQ}{KI}=\frac{AC}{KC}$. However, both fractions on the RHS are equal due to the external angle bisector theorem (similar to the internal angle bisector theorem). Thus $AP=AQ$, as desired.

To make a little change, we can have another problem: Let $ABC$ be a triangle with incenter $I$. Let $D$ be the point of intersection between the incircle and the side $BC$, the points $P$ is on the ray IB . Prove that: $\angle IAP=\angle CAD$ if and only if $PD \perp AD$.