Answer
See the detailed answer.
Work Step by Step
The uncetainity of the beam's location is in the vertical direction ($y$-axis).
So
$$\Delta p \Delta y\geq \hslash$$
Thus,
$$\Delta p_y \geq \dfrac{\hslash}{\Delta y}$$
where $\Delta y=D$
$$\Delta p_y \geq \dfrac{\hslash}{D}$$
And hence,
$$ p_y \approx \dfrac{\hslash}{D}\tag 1$$
Now let's work on the forward direction ($x$-direction) uncetainity of the beam's location .
By the same approach; but replacing $D$ with $\lambda$ since it is in the same direction of the wavelength.
$$ p_x\approx \dfrac{h}{\lambda}\tag 2$$
Therefore, the spread angle, for small angles, is given by
$$\phi=2\theta\approx 2\dfrac{p_y}{p_x}$$
Plugging from (1) and (2);
$$\phi \approx 2\dfrac{ \dfrac{\hslash}{D}}{ \dfrac{h}{\lambda}}$$
$$\phi \approx 2 \dfrac{ \hslash}{D}\dfrac{\lambda}{h}$$
$$\phi \approx \color{red}{\bf\not}2 \dfrac{ \color{red}{\bf\not}h}{\color{red}{\bf\not}2\pi D}\dfrac{\lambda}{\color{red}{\bf\not}h}$$
$$\phi \approx \dfrac{\lambda}{\pi D}$$
$$\boxed{\phi \approx\dfrac{\lambda}{ D}}$$
