Answer
See the detailed answer below.
Work Step by Step
We know that the potential at any point is the superposition of the potentials due to all charges.
So the potential at point $\rm p$ due to the $dq_{\rm ring}$, in the given figure, is given by
$$dV_{\rm ring}=\dfrac{1}{(4\pi \epsilon_0)}\dfrac{dq_{\rm ring}}{\sqrt{r^2+z^2}} \tag 1$$
Let's assume that the area charge density of this rod is $\sigma$ which is given by
$$\sigma=\dfrac{Q}{A}=\dfrac{Q}{\pi (R_{\rm out}^2-R_{\rm in}^2)}=\dfrac{dq_{\rm ring}}{2\pi rdr}$$
where $dq$ is the charge of the very thin black ring in the figure below and $dr$ is its width.
Thus,
$$dq_{\rm ring}=\dfrac{2\pi rQdr}{\pi (R_{\rm out}^2-R_{\rm in}^2)}$$
$$dq_{\rm ring}=\dfrac{2Qrdr}{(R_{\rm out}^2-R_{\rm in}^2)}\tag 2 $$
Plug $dq$ from (2) into (1),
$$dV_{\rm ring}=\dfrac{1}{(4\pi \epsilon_0)}\dfrac{ 1}{\sqrt{r^2+z^2}} \dfrac{2Qrdr}{(R_{\rm out}^2-R_{\rm in}^2)}$$
$$dV_{\rm ring}=\dfrac{2Q}{ (R_{\rm out}^2-R_{\rm in}^2) (4\pi \epsilon_0) }\dfrac{ rdr}{\sqrt{r^2+z^2}} $$
So the net electric potential at point $\rm p$ is given by
$$ V_{\rm disk}=\int dV_{\rm ring}=\dfrac{2Q}{ (R_{\rm out}^2-R_{\rm in}^2) (4\pi \epsilon_0) }\int_{R_{\rm in}}^{R_{\rm out}}\dfrac{ rdr}{\sqrt{r^2+z^2}} $$
$$ V_{\rm disk} =\dfrac{2Q}{ (R_{\rm out}^2-R_{\rm in}^2) (4\pi \epsilon_0) }\sqrt{r^2+z^2}\bigg|_{R_{\rm in}}^{R_{\rm out}} $$
$$\boxed{ V_{\rm disk} =\dfrac{2Q}{ (4\pi \epsilon_0)(R_{\rm out}^2-R_{\rm in}^2) }\left( \sqrt{R_{\rm out}^2+z^2}-\sqrt{R_{\rm in}^2+z^2}\right) }$$
