The De Broglie wavelength of a particle of mass $m$ and kinetic energy $K$ is:

For particles having the identical kinetic energy $K$:

Comparing the masses of the given particles:

• Electron: $m_e \approx 9.1 \times 10^{-31}\,\text{kg}$
• Proton: $m_p \approx 1\,\text{u}$
• Deuteron: $m_d \approx 2\,\text{u}$
• $\alpha$-particle: $m_\alpha \approx 4\,\text{u}$

Since the $\alpha$-particle has the largest mass, it possesses the shortest De Broglie wavelength. Hence, option 3 is the correct answer.