If we assume that penetrating power of any radiation/particle is inversely proportional to the De-broglie wavelength of the particle, then

$\lambda =\frac{h}{P}=\frac{h}{\sqrt{2mE}}\Rightarrow \frac{h}{\sqrt{2m(Vq)}}$

$\therefore$ For higher m and q ;
$\lambda$ will be smaller.
For an '$\alpha$' particle; both 'm' and 'q' are higher
hence lesser is the wavelength.
As, (penetrating power) $\propto$ Energy $\propto \frac{1}{\lambda }$
From above; penetrating power of an $\alpha -$particle is more than that of a proton.

So, the answer is option (B).