The de Broglie wavelengths of a proton and an α particle are λ and...

As we know, de Broglie wavelength is given by, $\lambda = \dfrac{h}{p} = \dfrac{h}{mv}$

$⇒ v = \dfrac{h}{m\lambda}$

Therefore, 

$\dfrac{{v}_{p}}{{v}_{\alpha}} = \dfrac{{m}_{\alpha}}{{m}_{p}} \times \dfrac{{\lambda}_{\alpha}}{{\lambda}_{p}}$

$= 4 \times 2 = 8$