A proton, an electron and an alpha particle have the same energies. Their de-Broglie wavelengths will be compared as :

For a particle of mass $$m$$ and kinetic energy $$E$$, the de-Broglie wavelength is given by

$$\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE}} \quad -(1)$$
since the linear momentum $$p = \sqrt{2mE}$$ for non-relativistic speeds.

From $$(1)$$ we get the proportionality

$$\lambda \propto \frac{1}{\sqrt{m}} \quad -(2)$$
for particles that possess the same energy $$E$$.

Therefore, for equal energies, the particle with the smaller mass has the larger wavelength.

Masses of the given particles:

Electron mass $$m_e \approx 9.11 \times 10^{-31}\,\text{kg}$$
Proton mass $$m_p \approx 1.67 \times 10^{-27}\,\text{kg}$$
Alpha-particle mass $$m_\alpha \approx 4m_p \approx 6.68 \times 10^{-27}\,\text{kg}$$

Hence

$$m_e \lt m_p \lt m_\alpha$$

Using $$(2)$$, the corresponding de-Broglie wavelengths satisfy

$$\lambda_e \gt \lambda_p \gt \lambda_\alpha$$

Re-ordering to match the option format:

$$\lambda_\alpha \lt \lambda_p \lt \lambda_e$$

This matches Option A.

Therefore, the correct comparison is
Case : $$\lambda_\alpha \lt \lambda_p \lt \lambda_e$$, which is Option A.