Answer
The magnitude of the net electric field at point P is $~~\frac{q}{4\pi ~\epsilon_0~d^2}~~$ and it points directly to the left.
Work Step by Step
We can see that by symmetry, all of the electric fields due to the charges cancel out, except for the $-2q$ on the left side of the large square and the $-q$ on the right side of the large square.
The net electric field at point P would equal the electric field due to a charge of $-q$ on the left side of the large square.
We can find the magnitude of this electric field:
$E = \frac{\vert -q \vert}{4\pi ~\epsilon_0~d^2} = \frac{q}{4\pi ~\epsilon_0~d^2}$
Note that an electric field points toward a negative charge.
Therefore, the magnitude of the net electric field at point P is $~~\frac{q}{4\pi ~\epsilon_0~d^2}~~$ and it points directly to the left.
