Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
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 dipole is given by
$$V_{\rm p}=V_1+V_2+V_3$$
$$V_{\rm p}=\dfrac{k_e(+q)}{r_1}+\dfrac{k_e(-2q)}{r_2}+\dfrac{k_e(+q)}{r_3}$$
From the figure below,
$$V_{\rm p}=\dfrac{k_eq}{y+s}-\dfrac{2k_eq}{y}+\dfrac{k_eq}{y-s}$$
$$V_{\rm p}=k_eq\left[\dfrac{1}{y+s}-\dfrac{ 2}{y }+\dfrac{1}{y-s}\right]$$
$$V_{\rm p}=k_eq\left[\dfrac{y(y-s)-2(y^2-s^2)+y(y+s)}{y(y^2-s^2)} \right]$$
$$V_{\rm p}=k_eq\left[\dfrac{y^2-ys-2y^2+2s^2+y^2+ys}{y(y^2-s^2)} \right]$$
$$V_{\rm p}=k_eq\left[\dfrac{ 2s^2 }{y(y^2-s^2)} \right]$$
and when $y\gt \gt s$, $[y^2-s^2\approx y^2]$, so
$$\boxed{V_{\rm p}= \dfrac{ 2k_eqs^2 }{y^3} }$$
$$\boxed{V_{\rm p}= \dfrac{1}{4\pi \epsilon_0}\dfrac{ 2 qs^2 }{y^3} }$$
