A simple pendulum of length L has a bob of mass m . It is released...

Let the velocity of the pendulum bob just before the collision be $u$. By conservation of mechanical energy, as it falls from a horizontal position (height $L$):
$u = \sqrt{2gL}$

The collision is head-on and elastic. The velocity $v_1$ of the bob of mass $m$ after colliding with a stationary block of mass $3m$ is:
$v_1 = \left( \dfrac{m - 3m}{m + 3m} \right) u$
$v_1 = \left( \dfrac{-2m}{4m} \right) u = -\dfrac{u}{2}$

The negative sign indicates that the bob rebounds. Its speed after the collision is $\dfrac{u}{2}$.

Let the maximum height reached by the bob during its rebound be $h$. Using conservation of mechanical energy again:
$\dfrac{1}{2}m v_1^2 = mgh$

Substituting $v_1 = \dfrac{u}{2}$ and $u^2 = 2gL$:
$\dfrac{1}{2}m \left( \dfrac{u}{2} \right)^2 = mgh$
$\dfrac{1}{2} \left( \dfrac{u^2}{4} \right) = gh$
$\dfrac{1}{2} \left( \dfrac{2gL}{4} \right) = gh$
$\dfrac{gL}{4} = gh$

$\Rightarrow h = \dfrac{L}{4}$

Answer: $\dfrac{L}{4}$