The kinetic energy of a body decreases by 36%. The decrease in its momentum is

$\mathrm{P}=\sqrt{2\mathrm{mE}} \therefore \mathrm{P}\propto \sqrt{\mathrm{E}}$

In given problem K.E. becomes 64% of the original value.
$\frac{{\mathrm{P}}_2}{{\mathrm{P}}_1}=\sqrt{\frac{{\mathrm{E}}_2}{{\mathrm{E}}_1}}=\sqrt{\frac{64\mathrm{E}}{100\mathrm{E}}}=0.8\Rightarrow {\mathrm{P}}_2=0.8\mathrm{P}$
$\therefore {\mathrm{P}}_2=80\mathrm{\%}$ of the original value.
i.e. decrease in momentum is 20%.