If the linear momentum is increased by 50%

Let ${P_1} = P$, ${P_2} = {P_1} + 50\% $ of ${P_1}$= ${P_1} + \frac{{{P_1}}}{2} = \frac{{3{P_1}}}{2}$ 

$E \propto {P^2}$

==> $\frac{{{E_2}}}{{{E_1}}} = {\left( {\frac{{{P_2}}}{{{P_1}}}} \right)^2} = {\left( {\frac{{3{P_1}/2}}{{{P_1}}}} \right)^2} = \frac{9}{4}$ 

==>${E_2} = 2.25E = {E_1} + 1.25{E_1}$

${E_2} = {E_1} + 125\% \;{\rm{of }}{E_1}$ 

i.e. kinetic energy will increase by $125\%.$