Given $m_Q=p{m}_{P }and v_Q=v_p/q.$From the principle
of conservation of momentum, we have (since body
P comes to rest after collision)

$m_P{v}_P+m_Q{v}_Q=m_{Q}v$

where v is the velocity of body Q after collision.
Thus

$m_P{v}_P+p{m}_P\times \frac{v_P}{q}=p{m}_{P}v.$

which gives            $\frac{v}{v_P}=\frac{p+q}{pq}$                       (i)

Now, the coefficient of restitution is given· by

$e=\frac{v}{v_P-v_Q}=\frac{v}{v_P-{\displaystyle \frac{v_P}{q}}}$

which gives             $\frac{v}{v_P}=\frac{e}{q}\left(q-1\right) \left(ii\right)$  

Equating (i) and (ii), we get  $e=\frac{p+q}{p\left(q-1\right)}$