If the kinetic energy of a moving body becomes four times its...

${K}_{2} = 4{K}_{1}$

$\dfrac{1}{2}m{v}_{2}^{2} = 4 \times \dfrac{1}{2}m{v}_{1}^{2}$

${v}_{2} = 2{v}_{1}$

$P = mv$

${P}_{2} = m{v}_{2} = 2m{v}_{1}$

${P}_{1} = m{v}_{1}$

$$ change $= \dfrac{\DeltaP}{{P}_{1}} \times 100 = \dfrac{2m{v}_{1} - m{v}_{1}}{m{v}_{1}} \times 100 = 100$