Explanation

(d) De Broglie wavelength, $\lambda = \frac{h}{p}$
Where $p$ is the momentum of the particle
For electron $\lambda_{e} = \frac{h}{p_{e}}$
For proton $\lambda_{p} = \frac{h}{p_{p}}$
As $\lambda_{e} = \lambda_{p}$ [Given]
$\Rightarrow p_{e} = p_{p}$
Or Momentum of electron = Momentum of proton