A particle of mass m is tied to a light string of length l having...

Let the fixed end of the string be the origin. The initial position of the particle is at a horizontal distance $l$ from the fixed end. The particle is projected upwards, so it travels a vertical distance $l$ to reach the highest point of the circular path.

For the particle to just complete the vertical circle, the tension in the string at the highest point must be exactly zero. This means the string momentarily becomes slack at the highest point.

At the highest point, the centripetal force is provided entirely by the weight of the particle:

$\dfrac{mv_H^2}{l} = mg \Rightarrow v_H = \sqrt{gl}$

Thus, the velocity at the highest point is $\sqrt{gl}$ and not zero.

Applying conservation of mechanical energy between the initial horizontal position and the highest point:

$\dfrac{1}{2}mv^2 = \dfrac{1}{2}mv_H^2 + mgl$

$\dfrac{1}{2}mv^2 = \dfrac{1}{2}m(gl) + mgl = \dfrac{3}{2}mgl$

The initial kinetic energy is $\dfrac{3}{2}mgl$, so the given statement for initial kinetic energy is incorrect.

Since the particle just completes the circle, it continues its circular motion and will pass through its initial position once again.