As per given figure, a weightless pulley $$P$$ is attached on a double inclined frictionless surface. The tension in the string (massless) will be (if $$g = 10 \text{ m s}^{-2}$$)

Lets write down equation for both

For 4 Kg

$$Mg\sin\theta_1\ -T=Ma$$

$$40\sin60^{\circ\ }\ -T=4a$$

For 1 Kg

$$T-mg\sin\theta_2\ =ma$$

$$T-10\sin30^{\circ\ }\ =a$$

Now add 2 equations

$$40\sin60^{\circ\ }-10\sin30^{\circ\ }\ =5a$$

$$40\times\ \frac{\sqrt{\ 3}}{2}-10\times\ \frac{1}{2}\ =5a$$

$$20\ \sqrt{\ 3}-5\ =5a$$

$$4\sqrt{\ 3}-1=a$$

Now substitute a in any equation

$$T-10\sin30^{\circ\ }\ =4\sqrt{\ 3}-1$$

$$T-5\ =4\sqrt{\ 3}-1$$

$$T=4\sqrt{\ 3}+4\ =\ 4\left(\sqrt{\ 3}+1\right)$$