A particle is projected at $60^{\circ }$ to the horizontal with a kinetic energy $K$. The kinetic energy at the highest point is:

At the point of projection, $K{E}_i=\frac{1}{2}m{v}^2=K$.In projectile motion, the horizontal component of velocity $v_x=v\cos \theta$ remains constant throughout the flight.At the highest point, the vertical component of velocity is zero. Thus, the velocity is $v_h=v\cos 60^{\circ }$.
$$v_h=v\left(\frac{1}{2}\right)=\frac{v}{2}$$
Kinetic energy at the highest point:
$$K{E}_h=\frac{1}{2}m{v}_h^2=\frac{1}{2}m{\left(\frac{v}{2}\right)}^2$$
$$K{E}_h=\frac{1}{4}\left(\frac{1}{2}m{v}^2\right)=\frac{K}{4}$$