b
From question,  \(m_{A}=M ; m_{B}=\frac{m}{2}\)

\(u_{A}=V \quad u_{B}=0\)

Let after collision velocity of \(A=V_{1}\) and velocity of \(B=V_{2}\) 

Applying law of conservation of momentum,

\(m u=m v_{1}+\left(\frac{m}{2}\right) v_{2}\)

or, \(24=2 \mathrm{v}_{1}+\mathrm{v}_{2}\)  ..... \((i)\)

By law of collision

\(e=\frac{v_{2}-v_{1}}{u-0}\)

or, \(u=v_{2}-v_{1}\)    ..... \((ii)\)        [ \(\because \) collision is elastic,  \(e=1\) ]

using eqns \((i)\) and \((ii)\)

\(\mathrm{v}_{1}=\frac{4}{3}\)  and  \(\mathrm{v}_{2}=\frac{4}{3} u\)

de-Broglie wavelength 

\(\lambda=\frac{h}{p}\)

\(\therefore \,\,\frac{{{\lambda _{\text{A}}}}}{{{\lambda _{\text{B}}}}} = \frac{{{P_B}}}{{{P_A}}} = \) \(\frac{{\frac{m}{2} \times \frac{4}{3}u}}{{m \times \frac{4}{3}}} = 2\)