Answer
$E$ is approximately proportional to $ \dfrac{1}{D^3}$ when $P$ is far away from the dipole.
Work Step by Step
We are given that $E=\dfrac{q}{D^2}-\dfrac{q}{(D+d)^2}=\dfrac{q}{D^2}(1-\dfrac{1}{(1+\dfrac{d}{D})^2})$
Re-arrange as follows:
$E=\dfrac{q}{D^2}-\dfrac{q}{(D+d)^2}=\dfrac{q}{D^2}(1-\dfrac{1}{(1+\dfrac{d}{D})^2})$
or, $E\approx \dfrac{q}{D^2}[1-(1-2(\dfrac{d}{D})+3(\dfrac{d}{D})^2-4(\dfrac{d}{D})^3)]$
$E\approx \dfrac{qd}{D^3}[2-3(\dfrac{d}{D})+4(\dfrac{d}{D})^2]$
It has been seen that when the point $P$ goes far away from the dipole, then the term $\dfrac{d}{D}$ becomes very small, which can be ignored.
Hence, we get $E\approx \dfrac{2qd}{D^3}$
This means that $E$ is approximately proportional to $\dfrac{1}{D^3}$ when $P$ is far away from the dipole.
