The exact electric field at point $P$ is

$\begin{align*}\mathbf{E} = \frac{q}{4 \pi \epsilon_0} \cdot \frac{r \hat{\mathbf{y}} - (l/2) \hat{\mathbf{x}} }{\left(r^2 + l...$
where the first term is from the positive charge and the second term is from the the negative charge. The y-components of the electric fields due to the two charges cancel, so
$\begin{align*}\mathbf{E} = -\frac{lq}{4 \pi \epsilon_0\left(r^2 + l^2/4 \right)^{3/2}} \hat{\mathbf{x}}\end{align*}$
When $r$ is much greater than $l$ , $\left(r^2 + l^2/4 \right)^{3/2} \approx r^3$ , so
$\begin{align*}\mathbf{E} = \boxed{-\frac{lq}{4 \pi \epsilon_0r^3} \hat{\mathbf{x}}}\end{align*}$
Thus, answer (E) is correct.

Solution to 2008 Problem 37