An electron and a proton having same momenta enter perpendicular into a magnetic field then

The radius of curvature of a charged particle of mass $m$ and charge $q$ moving in a region with uniform perpendicular field $B$ is given by, $r=\frac{m v}{B q}=\frac{p}{B q}$ where, $p$ is the momentum of the charged particle. Since momenta are same and masses for electron and proton are different, so they will attain different velocities and hence experience different forces. But the radius of circular path is depending on momenta so both will be moving on same trajectory (curved path, neglecting the sense of revolution).