Answer
$E_{net}=\frac{2kQa}{(x^2+a^2)^{3/2}}$ in the negative y-direction.
Work Step by Step
The horizontal x components of the two fields will cancel each other at the point P.
The net electric field is in the negative y direction. The charges are equal, so the net field is two times the y component of one of the electric field vectors.
$$E_{net}=2Esin\theta=2\frac{kQ}{x^2+a^2}sin\theta$$
$$E_{net}=2\frac{kQ}{x^2+a^2}\frac{a}{(x^2+a^2)^{1/2}}$$
$$E_{net}=\frac{2kQa}{(x^2+a^2)^{3/2}}$$
Again, this points in the negative y-direction.
