To find the ratio of velocities of two particles based on their de Broglie wavelengths, we can use the de Broglie wavelength formula, which relates the momentum of a particle to its wavelength. The de Broglie's wavelength formula is given by:

$$ \lambda = \frac{h}{p} $$

where:

$\lambda$ is the de Broglie wavelength,

$h$ is the Planck constant, and

$p$ is the momentum of the particle.

The momentum $p$ of a particle can also be expressed as the product of its mass $m$ and velocity $v$:

$$ p = mv $$

So, we can rewrite the de Broglie wavelength equation in terms of mass and velocity:

$$ \lambda = \frac{h}{mv} $$

For the proton (let's use subscript $p$ for proton), the wavelength is $\lambda$:

$$ \lambda_p = \frac{h}{m_p v_p}$$

For the $\alpha$ particle (let's use subscript $\alpha$ for alpha), the wavelength is $2\lambda$:

$$ 2\lambda = \frac{h}{m_\alpha v_\alpha}$$

We are interested in finding the ratio of the velocities $\frac{v_p}{v_\alpha}$. Using the given data about wavelengths:

$$ \lambda_p = \lambda $$

$$ \lambda_\alpha = 2\lambda $$

Using the de Broglie equation for both particles:

$$ \frac{h}{m_p v_p} = \lambda $$

$$ \frac{h}{m_\alpha v_\alpha} = 2\lambda $$

Dividing the second equation by the first equation gives us:

$$ \frac{\frac{h}{m_\alpha v_\alpha}}{\frac{h}{m_p v_p}} = \frac{2\lambda}{\lambda} $$

$$ \frac{m_p v_p}{m_\alpha v_\alpha} = \frac{2}{1} $$

$$ \frac{v_p}{v_\alpha} = \frac{2m_\alpha}{m_p} $$

Since we know an $\alpha$ particle consists of 2 protons and 2 neutrons (essentially four nucleons), the mass of an $\alpha$ particle is roughly four times the mass of a proton ($m_\alpha \approx 4m_p$).

Substituting $m_\alpha$ with $4m_p$ in the equation:

$$ \frac{v_p}{v_\alpha} = \frac{2(4m_p)}{m_p} $$

$$ \frac{v_p}{v_\alpha} = 2 \cdot 4 $$

$$ \frac{v_p}{v_\alpha} = 8 $$

Therefore, the ratio of the velocities of proton to $\alpha$ particle is $8:1$, which corresponds to Option A.