Answer
x = 2.41d.
Work Step by Step
The electric field strength drops off quickly. Any place where the field is zero is closer to the negative charge, since it is weaker. In between the two charges, the fields due to the two charges both point to the right, and cannot cancel.
At point P, the fields will cancel, so set the magnitudes equal.
$$k\frac{Q}{(d+x)^2}= k\frac{Q/2}{x^2}$$
$$2x^2=(x+d)^2$$
$$x=(\sqrt{2}+1)d\approx 2.41d$$