If momentum is increased by 20%, then K.E. increases by

$\mathrm{E}=\frac{{\mathrm{P}}^2}{2\mathrm{m}}.\text{ If }\mathrm{m}\text{ is constant then }\mathrm{E}\propto {\mathrm{P}}^2$

$\Rightarrow \frac{{\mathrm{E}}_2}{{\mathrm{E}}_1}={\left(\frac{{\mathrm{P}}_2}{{\mathrm{P}}_1}\right)}^2={\left(\frac{1.2\mathrm{P}}{\mathrm{P}}\right)}^2=1.44$

$\Rightarrow {\mathrm{E}}_2=1.44{\mathrm{E}}_1={\mathrm{E}}_1+0.44{\mathrm{E}}_1$

${\mathrm{E}}_2={\mathrm{E}}_1+44\mathrm{\%}\text{ of }{\mathrm{E}}_1$

i.e. the kinetic energy will increase by 44%