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

Let the point of suspension be $O$. The bob is released from a horizontal position (\theta = 90^{}^{{}^{\circ}}), so it falls through a vertical height $L$ before reaching its lowest point.

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

When the string is vertical, it hits the peg located at a distance $x$ below $O$. The bob then begins to move in a new vertical circle centered at the peg. The radius of this new circle is $R = L - x$.

For the bob to just complete a full vertical circle around the peg, the velocity at the lowest point of this new circle must be at least $\sqrt{5gR}$.

Therefore, we must have:
$v \ge \sqrt{5gR}$
$\sqrt{2gL} \ge \sqrt{5g(L - x)}$

Squaring both sides, we get:
$2gL \ge 5g(L - x)$
$2L \ge 5L - 5x$
$5x \ge 3L$
$\dfrac{x}{L} \ge \dfrac{3}{5}$
$\dfrac{x}{L} \ge 0.6$

Thus, the minimum value of $x/L$ required for the bob to execute a complete circle about the peg is $0.6$.