What Gives Equal Angles...?

The answer can easily be obtained from the dimensional analysis without doing any geometrical work.

However, here is an analytical geometric solution :

With the point $P$ on the circle , let the position coordinates of $A, B$ and $P$ be $(-a\cos α,a\sin α),(a\cos α,a\sin α)$ and $(a\cos β,a\sin β)$ respectively, where $a$ is the radius of the circle.

Then,

$|\overline {PR}|=2a\cos^2 \frac{α+β}{2}$ ,

$|\overline {PS}|=2a\sin^2 \frac{α-β}{2}$

$|\overline {PQ}|=a(\sin β-\sin α)$

$|\overline {PR}||\overline {PS}|=4a^2\cos^2 \frac{α+β}{2}\sin^2 \frac{α-β}{2}=a^2(\sin α-\sin β)^2$

Hence, $|\overline {PQ}|^2=|\overline {PR}||\overline {PS}|$

So, $k=\boxed 2$ .

Join $AP,BQ,QS,PB$ . As $\angle PRA +\angle PQA=\angle PSB +\angle PQB=180^\circ$ , we have $PQAR,PSBQ$ are cyclic quadrilaterals. So $\angle PRQ=\angle PAQ, \angle PQR=\angle PAR, \angle PSQ=\angle PBQ, \angle PQS =\angle PBS$ . So $\angle RAP =\frac{1}{2} arc {AP}=\angle PBA.$ So $\angle RQP=\angle RSQ$ and $\angle PRQ=\angle PQS$ . Hence, by AA criterion, $\triangle PRQ \sim \triangle PQS$ . Hence, $\frac{PR}{PQ}=\frac{PQ}{PS}$ so $PQ^2=PR \cdot PS$ . Hence, $k=2$ .