Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
As we see in the figure below, the potential at point $\rm P$ is given by
$$V=V_1+V_2=\dfrac{k_eq_1}{r_1}+\dfrac{k_eq_2}{r_2}$$
$$V= \dfrac{k_e q}{r_1}+\dfrac{-k_eq}{r_2}$$
where, from the geometry of the figure below, $r_1=\sqrt{x^2+\left(y-\dfrac{s}{2}\right)^2}$ and $r_2=\sqrt{x^2+\left(y+\dfrac{s}{2}\right)^2}$
$$V= k_e q\left[\dfrac{1}{\sqrt{x^2+\left(y-\dfrac{s}{2}\right)^2} }-\dfrac{1}{ \sqrt{x^2+\left(y+\dfrac{s}{2}\right)^2}}\right]$$
Recalling that $k_e=1/4\pi \epsilon_0$
$$\boxed{V=\dfrac{q}{4\pi \epsilon_0}\left[\dfrac{1}{\sqrt{x^2+\left(y-\dfrac{s}{2}\right)^2} }-\dfrac{1}{ \sqrt{x^2+\left(y+\dfrac{s}{2}\right)^2}}\right]}$$
$$\color{blue}{\bf [b]}$$
when $y\gt \gt s$ and $x\gt \gt s$, so we cna rewrite the boxed formula as follows to use the binomial approximiation as the author told us.
$$ V=\dfrac{q}{4\pi \epsilon_0}\left[\dfrac{1}{\sqrt{x^2+y^2\left(1-\dfrac{s}{2y}\right)^2} }-\dfrac{1}{ \sqrt{x^2+y^2\left( 1+\dfrac{s}{2y}\right)^2}}\right] $$
$$ V=\dfrac{q}{4\pi \epsilon_0}\left[\dfrac{1}{\sqrt{x^2+y^2\left(1-\dfrac{s}{y}+\dfrac{s^2}{4y^2}\right)} }-\dfrac{1}{ \sqrt{x^2+y^2\left( 1+\dfrac{s}{y}+\dfrac{s^2}{4y^2}\right)}}\right] $$
$$ V=\dfrac{q}{4\pi \epsilon_0}\left[\dfrac{1}{\sqrt{x^2+y^2 -ys+\dfrac{s^2}{4 } } }-\dfrac{1}{ \sqrt{x^2+y^2 +ys+\dfrac{s^2}{4 } }}\right] $$
$$ V=\dfrac{q}{4\pi \epsilon_0\sqrt{x^2+y^2}}\left[\dfrac{1}{\sqrt{ 1+\dfrac{-ys+\dfrac{s^2}{4 }}{x^2+y^2} } }-\dfrac{1}{ \sqrt{1+\dfrac{ +ys+\dfrac{s^2}{4 }}{x^2+y^2} }}\right] $$
$$ V=\dfrac{q}{4\pi \epsilon_0\sqrt{x^2+y^2}}\left[\left( 1+\dfrac{-ys+\dfrac{s^2}{4 }}{x^2+y^2} \right)^{-1/2}-\left(1+\dfrac{ +ys+\dfrac{s^2}{4 }}{x^2+y^2} \right)^{-1/2}\right] $$
Recaling that the binomial approximiation states that $(1+x)^n\approx 1+nx$ when $x\lt \lt 1$.
So $\left( 1+\dfrac{-ys+\dfrac{s^2}{4 }}{x^2+y^2} \right)^{-1/2}\approx 1-\dfrac{-ys+\dfrac{s^2}{4 }}{2(x^2+y^2)} $
and hence,
$\left(1+\dfrac{ +ys+\dfrac{s^2}{4 }}{x^2+y^2} \right)^{-1/2}\approx 1-\dfrac{ +ys+\dfrac{s^2}{4 }}{2(x^2+y^2)}$
Therefore,
$$ V=\dfrac{q}{4\pi \epsilon_0\sqrt{x^2+y^2}}\left[\left( 1-\dfrac{-ys+\dfrac{s^2}{4 }}{2(x^2+y^2)} \right) -\left(1-\dfrac{ +ys+\dfrac{s^2}{4 }}{2(x^2+y^2)} \right) \right] $$
$$ V=\dfrac{q}{4\pi \epsilon_0\sqrt{x^2+y^2}}\left[ -\dfrac{-ys+\dfrac{s^2}{4 }}{2(x^2+y^2)} +\dfrac{ ys+\dfrac{s^2}{4 }}{2(x^2+y^2)} \right] $$
$$ V=\dfrac{q}{4\pi \epsilon_0\sqrt{x^2+y^2}}\left[ \dfrac{ ys-\dfrac{s^2}{4 }}{2(x^2+y^2)} +\dfrac{ ys+\dfrac{s^2}{4 }}{2(x^2+y^2)} \right] $$
$$ V=\dfrac{q}{4\pi \epsilon_0\sqrt{x^2+y^2}}\left[ \dfrac{ ys}{ (x^2+y^2)} \right] $$
$$\boxed{ V=\dfrac{1}{4\pi \epsilon_0 }\left[ \dfrac{ qys}{ (x^2+y^2)^{3/2}} \right] }\tag 1$$
$$\color{blue}{\bf [c]}$$
We know that $E_s=-dV/ds$, so that
$$E_x=-\dfrac{d}{dx}V$$
Plug from (1),
$$E_x=-\dfrac{d}{dx}\dfrac{1}{4\pi \epsilon_0 }\left[ \dfrac{ qys}{ (x^2+y^2)^{3/2}} \right] $$
$$E_x=- \dfrac{1}{4\pi \epsilon_0 }\left[ \dfrac{ -3xqys(x^2+y^2)^{1/2}}{ (x^2+y^2)^3} \right] $$
$$\boxed{E_x= \dfrac{1}{4\pi \epsilon_0 }\left[ \dfrac{ 3xyqs}{ (x^2+y^2)^{5/2}} \right] }$$
And hence,
$$E_y=-\dfrac{d}{dy}V$$
$$E_y=-\dfrac{d}{dy}\dfrac{1}{4\pi \epsilon_0 }\left[ \dfrac{ qys}{ (x^2+y^2)^{3/2}} \right] $$
$$E_y=- \dfrac{1}{4\pi \epsilon_0 }\left[ \dfrac{ q s(x^2+y^2)^{3/2}-3qsy^2(x^2+y^2)^{1/2}}{ (x^2+y^2)^{3}} \right] $$
$$E_y= \dfrac{ qs}{4\pi \epsilon_0 }\left[ \dfrac{3 y^2 - (x^2+y^2) }{ (x^2+y^2)^{5/2}} \right] $$
$$\boxed{E_y= \dfrac{1}{4\pi \epsilon_0 }\left[ \dfrac{ qs(2 y^2 - x^2) }{ (x^2+y^2)^{5/2}} \right] }$$
$$\color{blue}{\bf [d]}$$
The on-axis means that $x=0$, so from the boxed formula of the electric field in $y$-axis in the $y$-axis,
$$E_y= \dfrac{1}{4\pi \epsilon_0 }\left[ \dfrac{ qs(2 y^2 - 0^2) }{ (0^2+y^2)^{5/2}} \right]$$
$$E_y= \dfrac{1}{4\pi \epsilon_0 }\left[ \dfrac{2qs y^2 }{ y^5} \right]$$
$$E_y= \dfrac{1}{4\pi \epsilon_0 }\left[ \dfrac{2qs }{ y^3} \right]$$
where $qs=p$
$$E_y= \dfrac{1}{4\pi \epsilon_0 }\left[ \dfrac{2p }{ y^3} \right]=E_{\rm dipole}\tag{on axis}$$
$$\color{blue}{\bf [e]}$$
The on the bisecting axis means that $y=0$, and hence $E_x=0$.
So from the boxed formula of the electric field in $y$-axis in the $y$-axis,
$$ E_y= \dfrac{1}{4\pi \epsilon_0 }\left[ \dfrac{ qs(0^2 - x^2) }{ (x^2+0^2)^{5/2}} \right] $$
$$ E_y= \dfrac{1}{4\pi \epsilon_0 }\left[ \dfrac{ -qs x^2 }{ x^5} \right] $$
$$ E_y= \dfrac{1}{4\pi \epsilon_0 }\left[ \dfrac{ -p}{ x^3} \right] =E_{\rm dipole}\tag{bisecting axis}$$
