As per Angle Bisector theorem,
$\qquad\;\;\dfrac{QS}{SR} =\dfrac{r}{q}$
$\dfrac{QS}{(p-QS)} =\dfrac{r}{q}$
$\qquad \quad QS =\dfrac{pr}{(q+r)} \quad \quad \longrightarrow(1)$
We have in a triangle $\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}$
So, from $\triangle QPS,\;\; \dfrac{QS}{\sin 60} =\dfrac{PS}{\sin Q}$
$\qquad PS =\dfrac{QS\times \sin Q}{\sin 60}\quad \quad\longrightarrow (2)$
From $\triangle PQR,\;\; \dfrac{p}{\sin 120} =\dfrac{q}{\sin Q}$
$p =\dfrac{q\times \sin 120}{\sin Q} =\dfrac{q\times \sin 60}{\sin Q}\quad \quad\longrightarrow (3)$
So, from $(1), (2)$ and $(3),$
$PS =\dfrac{qr}{(q+r)}$
B choice.
http://en.wikipedia.org/wiki/Angle_bisector_theorem