Answer
$E= -kQ\frac{4xa}{(x^2-a^2)^2}$.
Work Step by Step
Let the rightward direction be positive. The field due to +Q is positive, while the field due to -Q is negative. The negative charge is closer to point P so we expect the net field to point to the left.
$$E=k\frac{Q}{(x+a)^2}+ k\frac{-Q}{(x-a)^2}$$
$$E=kQ(\frac{1}{(x+a)^2}-\frac{1}{(x-a)^2})$$
$$E= -kQ\frac{4xa}{(x^2-a^2)^2}$$
As expected, the field points to the left.
The magnitude $E= \frac{4kQxa}{(x^2-a^2)^2}$.