1. Magnetic force on a charged particle

   When a charged particle enters a magnetic field perpendicular to it, the magnetic force is $F = qvB$ and this force acts as the centripetal force: $qvB = \frac{mv^2}{r}$

2. Radius of circular path

   From the above, $r = \frac{mv}{qB}$

   Since momentum $p = mv$, we can write $r = \frac{p}{qB}$

3. Compare electron and proton

   The electron and proton have the same momentum and both enter perpendicular to the magnetic field.

   Their charges in magnitude are equal: $|q_e| = |q_p| = e$

   Therefore, their radii are $r_e = \frac{p}{eB}, \qquad r_p = \frac{p}{eB}$

   So, $r_e = r_p$

4. Effect of sign of charge

   Electron has charge $-e$ and proton has charge $+e$, so they bend in opposite senses. But the question says to ignore the sense of revolution.

   Hence, their curved paths have the same curvature.

5. Check options

   • A: curved path of electron and proton will be same (ignoring the sense of revolution) — Correct
   • B: they will move undeflected — Incorrect, because magnetic force is nonzero for motion perpendicular to $\vec B$
   • C: curved path of electron is more curved — Incorrect
   • D: path of proton is more curved — Incorrect

Final Answer: $\boxed{\text{A}}$