The electric field at a point on the equatorial plane at a distance...

Use the concept of potential gradient, then the relation, ${E}_{r} = - \dfrac{d V}{d r} ⇒ {E}_{\theta} = - \dfrac{1}{r} \times \dfrac{d V}{d \theta} \text{ and } V = \dfrac{KP}{{r}^{2}}cos\theta$, for general point,
$⇒ {E}_{\theta} = - \dfrac{1}{r} \times \dfrac{d \left(\dfrac{KP}{{r}^{2}}cos\theta\right)}{d \theta} ⇒ E = - \dfrac{KP}{{r}^{3}}\sqrt{1 + 3\cos^{2}\theta}$, for the equatorial position, $\theta = 90 °$

$⇒ \overset{\to}{E} = - \dfrac{k\overset{\to}{P}}{{r}^{3}}$$= - \overset{\to}{P}4\pi{ϵ}_{0}{r}^{3}$.