de Broglie wavelength: $\lambda=\dfrac{h}{p}=\dfrac{h}{\sqrt{2mK}}$ (in terms of KE) or $\lambda=\dfrac{h}{\sqrt{2mqV}}$ (accelerated through potential $V$).

Taking $m_\alpha=4m_p$ and $q_\alpha=2e$, $q_p=e$:

(i) Same kinetic energy $K$:

$$\frac{\lambda_\alpha}{\lambda_p}=\sqrt{\frac{m_p}{m_\alpha}}=\sqrt{\frac{m_p}{4m_p}}=\frac12$$

(ii) Same accelerating potential $V$:

$$\frac{\lambda_\alpha}{\lambda_p}=\sqrt{\frac{m_pq_p}{m_\alpha q_\alpha}}=\sqrt{\frac{m_p\times e}{4m_p\times2e}}=\sqrt{\frac18}=\frac{1}{2\sqrt2}