The field between the plates is $E = U/d$, and on a positive $\alpha$ the force is from $A$ to $B$. Take the emission point as the origin, with the $x$-axis from $A$ to $B$. An $\alpha$-particle with initial speed $v_0 \le v$ and emission angle $\theta$ from the $x$-axis has:

$$x(t) = v_0 \cos\theta\, t + \tfrac{1}{2} a t^2, \qquad y(t) = v_0 \sin\theta\, t,$$

with $a = qE/m = qU/(md)$.

For a fixed $v_0$, the lateral displacement on $B$ (set $x = d$) is largest when $\theta = 90^\circ$ (perpendicular emission): increasing $v_x(0)$ shortens the flight time more than it adds to nothing in $y$. So the maximum lateral hit on $B$ is achieved by the fastest particle emitted perpendicular to the field. For that particle the flight time is

$$t_{\max} = \sqrt{\dfrac{2d}{a}} = d \sqrt{\dfrac{2m}{qU}},$$

and the lateral displacement is

$$r_{\max} = v \, t_{\max} = v d \sqrt{\dfrac{2m}{qU}}.$$

By azimuthal symmetry around the field direction, every point with $r \le r_{\max}$ is hit by some $\alpha$-particle (smaller $v_0$ or non-perpendicular emission fills in smaller radii). So the fluorescent region is a filled disk of radius $r_{\max}$, with area

$$S = \pi r_{\max}^2 = \pi v^2 d^2 \cdot \dfrac{2m}{qU} = \dfrac{2\pi m v^2 d^2}{qU}.$$