Let ball strikes at a speed $u$ the $K_1=\frac{1}{2}m{u}^2$

Due to collision tangential component of velocity remains unchanged at $u$ sin 45$°$, but the normal component of velocity change to
$u$ sin45$°=\frac{1}{2}u\cos 45°$

$\therefore$ Final velocity of ball after collision

$v={\sqrt{{\left(u\sin 45°\right)}^2+{\left(\frac{1}{2}u\cos 45°\right)}^2}}^{ }$

$=\sqrt{{\left(\frac{u}{\sqrt{2}}\right)}^2+{\left(\frac{u}{2\sqrt{2}}\right)}^2}=\sqrt{\frac{5}{3}}u.$

Hence final kinetic energy $K_2=\frac{1}{2}m{v}^2=\frac{5}{16}m{u}^2$.

$\therefore$ Fractional loss in KE

$=\frac{K_1-K_2}{K_1}=\frac{\frac{1}{2}m{u}^2-\frac{5}{16}m{u}^2}{\frac{1}{2}m{u}^2}=\frac{3}{8}$