Friction plays its role

See Third equation of motion

Use $\small{s=\dfrac{v^2-u^2}{2a}}$ .

$\dfrac mn=\dfrac{\frac{v^2}{2g\sin \alpha}}{\frac{-v^2}{2g(\sin \alpha-\mu \cos\alpha)}}$

( Note:- Final velocity on smooth portion is same as initial velocity on rough portion)

$\implies \dfrac mn=\dfrac{\mu \cos \alpha-\sin \alpha}{\sin \alpha}=\mu\cot \alpha-1$

$\mu=\boxed{\left(\dfrac{m+n}{n}\right)\tan \alpha}$ .

The potential energy of particle at top of plane is $Mg{\space}(m+n)k{\space}sin\alpha$

Now this potential energy is totally used into frictional energy for distance $nk$ .

So frictional energy is $\mu {\space}Mg{\space}nk{\space}cos\alpha$ $=Mg{\space}(m+n)k{\space}sin\alpha$

Therefore we got $\mu=\dfrac{[m+n]*tan\alpha}{n}$