Let the point $P$ lies at a distance $x$ from the centre of the disk and take the plane of the disk to be perpendicular to the $\mathbf{x}$-axis. Let the disc is divided into a number of charged rings as shown in figure.

The electric potential of each ring, of radius $r$ and width $d r$, have charge $d q$ is given by

$$\sigma d A=\sigma 2 \pi r d r$$

and potential is given by

$$d V=\frac{k_e d q}{\sqrt{r^2+x^2}}=\frac{k_e \sigma 2 \pi r d r}{\sqrt{r^2+x^2}}$$

where $k_e=\frac{1}{4 \pi \varepsilon_0}$ the total electric potential at $P$, is given by

$$\begin{aligned} &\begin{aligned} & V=\pi k_e \sigma \int_0^a \frac{2 r d r}{\sqrt{r^2+x^2}}=\pi k_e \sigma \int_0^a\left(r^2+x^2\right)^{-1 / 2} 2 r d r \\ & V=2 \pi k_e \sigma\left[\left(x^2+a^2\right)^{1 / 2}-x\right] \end{aligned}\\ &\begin{aligned} \text{So, we have by substring}\quad k_e & =\frac{1}{4 \pi \varepsilon_0} \\ V & =\frac{1}{4 \pi \varepsilon_0} \frac{2 Q}{a^2}\left[\sqrt{x^2+a^2}-x\right] \end{aligned} \end{aligned}$$

Let any arbitrary point on the required plane is $(x, y, z)$. The two charges lies on $z$-axis at a separation of $2 d$.

The potential at the point $P$ due to two charges is given by

$$\begin{array}{ll} & \frac{q_1}{\sqrt{x^2+y^2+(z-d)^2}}+\frac{q_2}{\sqrt{x^2+y^2+(z+d)^2}}=0 \\ \therefore & \frac{q_1}{\sqrt{x^2+y^2+(z-d)^2}}=\frac{q_2}{\sqrt{x^2+y^2+(z+d)^2}} \end{array}$$

On squaring and simplifying, we get

$$x^2+y^2+z^2+\left[\frac{\left(q_1 / q_2\right)^2+1}{\left(q_1 / q_2\right)^2-1}\right](2 z d)+d^2-0$$

This is the equation of a sphere with centre at

$$\left(0,0,-2 d\left[\frac{q_1^2+q_2^2}{q_1^2-q_2^2}\right]\right)$$

Let third charge $+q$ is slightly displaced from mean position towards first charge. So, the total potential energy of the system is given by

$$\begin{aligned} U & =\frac{1}{4 \pi \varepsilon_0}\left\{\frac{-q^2}{(d-x)}+\frac{-q^2}{(d+x)}\right\} \\ U & =\frac{-q^2}{4 \pi \varepsilon_0} \frac{2 d}{\left(d^2-x^2\right)} \\ \frac{d U}{d x} & =\frac{-q^2 \cdot 2 d}{4 \pi \varepsilon_0} \cdot \frac{2 x}{\left(d^2-x^2\right)^2} \end{aligned}$$

The system will be in equilibrium, if

$$F=-\frac{d U}{d x}=0$$

On solving, $x=0$. So for, $+q$ charge to be in stable/unstable equilibrium, finding second derivative of PE.

$$\begin{aligned} \frac{d^2 U}{d x^2} & =\left(\frac{-2 d q^2}{4 \pi \varepsilon_0}\right)\left[\frac{2}{\left(d^2-x^2\right)^2}-\frac{8 x^2}{\left(d^2-x^2\right)^3}\right] \\ & =\left(\frac{-2 d q^2}{4 \pi \varepsilon_0}\right) \frac{1}{\left(d^2-x^2\right)^3}\left[2\left(d^2-x^2\right)^2-8 x^2\right] \\ \text{At}\quad x & =0 \\ \frac{d^2 U}{d x^2} & =\left(\frac{-2 d q^2}{4 \pi \varepsilon_0}\right)\left(\frac{1}{d^6}\right)\left(2 d^2\right), \text { which is }<0 \end{aligned}$$

This shows that system will be unstable equilibrium.