The kinetic energy is given by $KE =\frac{p^{2}}{2 m}$
So, $\Delta KE =\frac{2 p \Delta p}{2 m}=\frac{p \Delta p}{m}$
$\Rightarrow \frac{\Delta KE }{ KE }=\frac{p \Delta p}{m} \times \frac{2 m}{p^{2}}=\frac{2 \Delta p}{p}$
Since momentum $p$ increases by $20 \%$, so the final momentum becomes $1.2 \,p$.
Hence, $K E_{\text {final }}=\frac{(1.2 p)^{2}}{2 m}=1.44 \frac{p^{2}}{2 m}=1.44 \,KE$
So, $\%$ change in $KE =44 \%$.