Let the point $P$ be at the distance $r$ from the centre of dipole on the side of the charge $q$, then

 Here $\displaystyle E_{-q} =\frac{kq}{( r+a)^{2}}$   and    $\displaystyle E_{+q} =\frac{kq}{( r- a)^{2}}$ 

So the total electric field at pont $\displaystyle P$ is

$\displaystyle E=E_{+q} -E_{-q}$                       $\displaystyle \ ( E_{+q}  >E_{-q})$

$\displaystyle E=kq\left[\frac{kq}{( r- a)^{2}} -\frac{kq}{( r+a)^{2}}\right]$ 

$\displaystyle E=kq\cdotp \frac{4ar}{\left( r^{2} -a^{2}\right)^{2}}$

For $\displaystyle ( r\ggg a)$          $\displaystyle E=\frac{4kqa}{r^{3}}$   $\displaystyle \vec{E} =\frac{2k\vec{P}}{r^{3}}$

The direction of $\displaystyle \vec{E}$ is along $\displaystyle \vec{P}$ i.e from $\displaystyle -q$ to $\displaystyle +q$

#$\displaystyle E\varpropto \ \frac{1}{r^{3}}$