1. Maximum kinetic energy of emitted electron

For photoelectric emission, $K_{\max}=\frac{hc}{\lambda}-\phi$

Let $K_{\max}=\frac{1}{2}mv^2$

So, $\frac{1}{2}mv^2=\frac{hc}{\lambda}-\phi$

Hence, $v=\sqrt{\frac{2}{m}\left(\frac{hc}{\lambda}-\phi\right)}$

2. Motion in magnetic field

The magnetic field is perpendicular to the initial velocity, so the electron moves in a circular path of radius $r=\frac{mv}{eB}$

Substitute the value of $v$: $r=\frac{m}{eB}\sqrt{\frac{2}{m}\left(\frac{hc}{\lambda}-\phi\right)}$

$r=\frac{\sqrt{2m\left(\frac{hc}{\lambda}-\phi\right)}}{eB}$

3. Where does the electron hit the plate again?

The electron is emitted normally from the plate, so its initial velocity is perpendicular to the plate.

Since the magnetic field exists throughout the path and is perpendicular to velocity, the electron follows a circle. Starting from the plate with velocity normal to it, the electron returns to the plate after completing a semicircle.

Therefore, the distance between the emission point $A$ and the return point $B$ is the diameter of the circular path: $AB=2r$

So, $AB=2\cdot \frac{\sqrt{2m\left(\frac{hc}{\lambda}-\phi\right)}}{eB}$

$AB=\frac{\sqrt{8m\left(\frac{hc}{\lambda}-\phi\right)}}{eB}$

4. Match with options

This corresponds to: $\boxed{\frac{\sqrt{8m\left(\frac{hc}{\lambda}-\phi\right)}}{eB}}$

So the correct option is B.