Solution:

First note that $ID = IE = IF$ because they are all radii of the incircle, and $\angle BFI = \angle BDI = 90^{\circ}$ because tangents are perpendicular to radii. Since $\angle ABC = 90^{\circ}$ we have $BFID$ a square and so $BD = BF = ID$ too. Thus $\triangle EIF$ is isosceles and so $\angle IFE = \angle FEI$.

Since $CI$ is the bisector of $\angle DCE$, we see that $D$ and $E$ are reflections of each other over line $CIP$. Therefore $\angle IDP = \angle PEI$. Hence
$$\angle IDP = \angle FEI = \angle IFP$$
and therefore quadrilateral $FPID$ is cyclic. Therefore $\angle FPD = \angle FID = 90^{\circ}$ ($BFID$ is a square). Since $AI$ is the bisector of $\angle EAF$, we see that $E$ and $F$ are reflections of each other over line $AI$. Therefore $EF \perp AI$. Hence
$$AI \parallel DQ$$
because they are both perpendicular to $EF$. We also have $AQ \parallel ID$ (because $BFID$ is a square) so $QAID$ is a parallelogram. Therefore
$$AQ = ID = BF$$
as required.