The de-Broglie hypothesis ties wavelength solely to momentum, not to mass, speed, or energy directly. Since $\lambda = h/p$, fixing $\lambda$ fixes $p$ for any particle.
Given electron and proton share $\lambda = 0.2\ \text{\AA}$, both carry the identical momentum $p = h/\lambda$. That immediately makes option (C) correct.
Now rule out the rest using the mass difference. Speed follows $v = p/m$; the proton being roughly 1836 times heavier moves about 1836 times slower, so speeds are not equal, eliminating (B).
Kinetic energy for equal momentum is $K = p^2/(2m)$, which is inversely proportional to mass. So the electron, being far lighter, has the much larger kinetic energy, and the proton has the smaller one. This kills both (A) and (D), the latter claiming the proton has more energy.
Hence the only shared quantity is momentum.
\[\boxed{\text{Both have the same momentum } p = h/\lambda}\]