A simple pendulum of length L with a bob of mass m is deflected...

Let the lowest point of the pendulum be the reference level for potential energy. The initial height of the bob when released from \theta = 90^{}^{{}^{\circ}} is $L$.

By conservation of mechanical energy, the velocity $v_0$ of the bob at the lowest point is given by:
$\dfrac{1}{2} m v_0^2 = mgL \Rightarrow v_0 = \sqrt{2gL}$

When the string strikes the peg at a distance $x = L/2$ below the suspension point, the bob starts moving in a new circle of radius $R = L - x = L/2$ centered at the peg.

Let the string become slack when it makes an angle $\theta$ with the upward vertical. The height $h$ of the bob above the lowest point is:
$h = R + R \cos\theta = \dfrac{L}{2}(1 + \cos\theta)$

Using conservation of energy to find the velocity $v$ at this height:
$\dfrac{1}{2} m v^2 = mgL - mgh$
$\dfrac{1}{2} m v^2 = mgL - mg\dfrac{L}{2}(1 + \cos\theta) = mg\dfrac{L}{2}(1 - \cos\theta)$
$v^2 = gL(1 - \cos\theta)$

The tension $T$ in the string at this position is provided by the centripetal force equation:
$T + mg \cos\theta = \dfrac{mv^2}{R}$
$T = \dfrac{m(gL(1 - \cos\theta))}{L/2} - mg \cos\theta$
$T = 2mg(1 - \cos\theta) - mg \cos\theta$
$T = mg(2 - 3\cos\theta)$

The string becomes slack when the tension $T = 0$:
$mg(2 - 3\cos\theta) = 0 \Rightarrow \cos\theta = \dfrac{2}{3}$

The height of the bob above its lowest position at the instant the string becomes slack is:
$h = \dfrac{L}{2}\left(1 + \dfrac{2}{3}\right) = \dfrac{L}{2} \times \dfrac{5}{3} = \dfrac{5L}{6}$

Answer: $5L/6$