If the kinetic energy of the particle is increased to 16 times its previous value, the percentage change in…

$K{E}_{final}=16K{E}_{initial}$

$\Rightarrow \frac{1}{2}m{v_f}^2=16\times \frac{1}{2}m{v_i}^2$(mass remains same because the particle remains same)

$\Rightarrow v_f=4v_i$
We know that, De-broglie wavelength $\left(\lambda \right)=\frac{h}{mv}$

$\therefore {\lambda }_f=\frac{h}{m{v}_f}=\frac{h}{4m{v}_i}$

  So De-Broglie wave will decreased to $75\%$