In simple harmonic, force on charge $\mathrm{q}$ in motion must be proportional to its distance from the centre $\mathrm{O}$ and is directed towards $\mathrm{O}$.
As given that, two charge $-\mathrm{q}$ at $\mathrm{A}$ and $\mathrm{B}$
$$
\mathrm{AB}=[\mathrm{AO}+\mathrm{OB}]=2 \mathrm{~d}(\because \mathrm{AO}=\mathrm{OB}=\mathrm{d})
$$
$\mathrm{x}$ is small distance perpendicular to $\mathrm{O}$, i.e.
$\mathrm{x} < \mathrm{d}$ mass of charge $\mathrm{q}$.

So, force of attraction at $P$ towards $A$ and $B$ are each
$\mathrm{F}=\frac{\mathrm{q}(\mathrm{q})}{4 \pi \varepsilon_0 \mathrm{r}^2}$, whereAP $=\mathrm{BP}=\mathrm{r}$
Horizontal components of these forces $\mathrm{F}_{\mathrm{n}}$ are cancel out. Vertical components along PO add.
If $\angle A P O=O$, the net force on $q$ along $P O$ is
$$
\begin{aligned}
F^{\prime} &=2 F \cos Q \\
&=\frac{2 q^2}{4 \pi \varepsilon_0 r^2}\left(\frac{x}{r}\right)=\frac{2 q^2 x}{4 \pi \varepsilon_0\left(d^2+x^2\right)^{3 / 2}}
\end{aligned}
$$
When, $x< < d, F^{\prime}=\frac{2 q^2 x}{4 \pi \varepsilon_0 d^3}=K x$
where $\quad K=\frac{2 q^2}{4 \pi \varepsilon_0 d^3}$
So, $\quad F \propto x$
Thus, force on charge $\mathrm{q}$ is proportional to its displacement from the centre $\mathrm{O}$ and it is directed towards $\mathrm{O}$.
Hence, motion of charge $q$ would be simple harmonic, where
$$
\omega=\sqrt{\frac{\mathrm{K}}{\mathrm{m}}}
$$
and $\mathrm{T}=\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{K}}}=2 \pi \sqrt{\frac{\mathrm{m} 4 \pi \varepsilon_0 \mathrm{~d}^3}{2 \mathrm{q}^2}}$
$$
\left[\because \mathrm{K}=\frac{2 \mathrm{q}^2}{4 \pi \varepsilon_0 \mathrm{~d}^3}\right] \quad\left[\mathrm{T}=\frac{8 \pi^3 \varepsilon_0 \mathrm{md}^3}{\mathrm{q}^2}\right]^{1 / 2}
$$