As per Einstein's photoelectric equation, $\frac{hc}{\lambda} = \phi_0 + k$. Given $\phi_0 \to 0$, so $\frac{hc}{\lambda} = k = \frac{P^2}{2m} \Rightarrow P = \sqrt{\frac{2mhc}{\lambda}}$. Now De-broglie wavelength, $\lambda_d = \frac{h}{P} = \frac{h}{\sqrt{2mhc/\lambda}} \Rightarrow \sqrt{\lambda} = \lambda_d \sqrt{\frac{2mc}{h}} \Rightarrow \lambda = \left(\frac{2mc}{h}\right) \lambda_d^2$.