The De Broglie wavelength is $\lambda =\frac{\mathrm{h}}{\mathrm{mv}}$

The momentum is given by $p=mv$

The momentum of proton is $p_p=m_p{v}_p$

The momentum of electron is  $p\text{'}=m_e{v}_e$

Since $h$ is the Planck's constant and it is given that both wavelengths are same, i.e.

${\lambda }_p={\lambda }_e$, so $p_p=p\text{'}$.