All the surfaces and pulleys are frictionless in the shown arrangement. Pulleys P and Q are massless. The force applied by clamp on pulley P is 

$$\begin{gathered} a=\frac{\text{ Net pulling force }}{\text{ Total mass }} \\ =\frac{2mg\sin 30°}{2m+m}=\frac{g}{3} \end{gathered}$$

FBD of m

$T=m a=m g / 3$

Resultant of tensions

$\mathbf{R}=-\left(T\cos 30°\right)\hat{\mathbf{i}}-\left(T\sin 30°+T\right)\hat{\mathbf{j}}$

Putting$T=mg/3$

$\mathbf{R}=-\frac{\sqrt{3}}{6}mg\hat{\mathbf{i}}-\frac{mg}{2}\hat{\mathbf{j}}$

Since, pulley P is in equilibrium. Therefore,

$\mathbf{F}+\mathbf{R}=0$

where, $\mathbf{F}=$ force applied by clamp on pulley

$\therefore \mathbf{F}=-\mathbf{R}=\frac{mg}{6}(\sqrt{3}\hat{\mathbf{i}}+3\hat{\mathbf{j}})$