Answer
a) $ \dfrac{1}{4\pi \epsilon_0} \dfrac{ p}{ { y^2 } } $
b) $0.056\;\rm V$
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
$$V_{\rm p}=V_++V_-$$
$$V_{\rm p}=\dfrac{k_e(+q)}{r_+}+\dfrac{k_e(-q)}{r_-}$$
From the figure below,
$$V_{\rm p}=\dfrac{k_eq}{y-\dfrac{s}{2}}-\dfrac{k_eq}{y+\dfrac{s}{2}}$$
$$V_{\rm p}=k_eq\left[\dfrac{1}{y-\dfrac{s}{2}}-\dfrac{ 1}{y+\dfrac{s}{2}}\right]$$
$$V_{\rm p}=k_eq\left[\dfrac{1}{ \dfrac{2y-s}{2}}-\dfrac{ 1}{\dfrac{2y+s}{2}}\right]$$
$$V_{\rm p}=2k_eq\left[\dfrac{1}{ {2y-s} }-\dfrac{ 1}{ {2y+s} }\right]$$
$$V_{\rm p}=2k_eq\left[\dfrac{2y+s-[2y-s]}{ {(2y-s)(2y+s)} } \right]$$
$$V_{\rm p}=2k_eq\left[\dfrac{2s}{ {(2y-s)(2y+s)} } \right]$$
$$V_{\rm p}=4k_eqs\left[\dfrac{1}{ { 4y^2-s^2 } } \right]$$
$$V_{\rm p}= \dfrac{4k_ep}{ { 4y^2-s^2 } } $$
$$V_{\rm p}= \dfrac{ k_ep}{ { y^2-\dfrac{s^2}{4} } } $$
and since $y\gt\gt s$, so $y^2-\dfrac{s^2}{4} \approx y^2$
$$\boxed{V_{\rm p}= \dfrac{ k_ep}{ { y^2 } }=\dfrac{1}{4\pi \epsilon_0} \dfrac{ p}{ { y^2 } } }$$
$$\color{blue}{\bf [b]}$$
Plug the known into the boxed formula above,
$$ V_{\rm p}= \dfrac{1}{4\pi \epsilon_0} \dfrac{ p}{ { y^2 } }=\dfrac{ (9\times 10^9)(6.2\times 10^{-30})}{ { (1.0\times10^{-9})^2 } }$$
$$ V_{\rm p}=\color{red}{\bf 0.056}\;\rm V$$
