Trig Problem

We know $AD=BA=3+2=5$ 

$AQ=AP= \sqrt{AD^2+DQ^2}=\sqrt{34}$

$\triangle AQD \cong \triangle ABP$ by SSS congruence
Thus:
$\angle QAD=\angle PAB$ (by congruence above)
And $\angle PAQ=90-2\angle PAB$ 
 Let $\angle PAB =x$and $\angle PAQ=90-2x$    (for simplicity)
 

$\sin(90-2x)=\cos(2x)=1-2\sin^2x$

We know $\sin x=\frac {PB}{AP}=\frac{3}{\sqrt{34}}$
so finally:

$\sin {\angle PAQ}=\boxed{\frac{8}{17}}$
ggwp