A particle of mass m moving with velocity v collides completely...

Using the principle of conservation of linear momentum:

Initial momentum = Final momentum
$m v + (3m)(0) = (m + 3m) v_f$
$m v = 4m v_f$
$v_f = \dfrac{v}{4}$

Initial kinetic energy of the system:
$K_i = \dfrac{1}{2} m v^2$

Final kinetic energy of the combined system:
$K_f = \dfrac{1}{2} (4m) v_f^2 = \dfrac{1}{2} (4m) \left(\dfrac{v}{4}\right)^2 = \dfrac{1}{2} m v^2 \left(\dfrac{1}{4}\right) = \dfrac{K_i}{4}$

Loss in kinetic energy:
$\Delta K = K_i - K_f = K_i - \dfrac{K_i}{4} = \dfrac{3K_i}{4}$

Percentage loss in kinetic energy:
$\dfrac{\Delta K}{K_i} \times 100 = \dfrac{3}{4} \times 100 = 75\%$

Answer: $75\%$