Answer
$E$ is approximately proportional to $ \dfrac{1}{D^3}$ when $P$ is far away from the dipole.
Work Step by Step
$E=\dfrac{q}{D^2}-\dfrac{q}{(D+d)^2}$
Re-write as: $E=\dfrac{q}{D^2}(1-\dfrac{1}{(1+\dfrac{d}{D})^2})$
Use the Binomial series.
we get
$E\approx \dfrac{q}{D^2}[1-(1-2(\dfrac{d}{D})+3(\dfrac{d}{D})^2-4(\dfrac{d}{D})^3)]$
$E\approx \dfrac{qd}{D^3}[2-3(\dfrac{d}{D})+4(\dfrac{d}{D})^2]$
It has been shown that when the point $P$ is far away from the dipole, then the term $\dfrac{d}{D}$ becomes very small and can be neglected.
Therefore, we have $E\approx \dfrac{2qd}{D^3}$
Hence, it has been proved that $E$ is approximately proportional to $ \dfrac{1}{D^3}$ when $P$ is far away from the dipole.
