Proton ( P ) and electron ( e ) will have same de-Broglie...

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

The momentum is given by $p = mv$

The momentum of proton is ${p}_{p} = {m}_{p}{v}_{p}$

The momentum of electron is $p ' = {m}_{e}{v}_{e}$

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

${\lambda}_{p} = {\lambda}_{e}$, so ${p}_{p} = p '$.