Forces acting on charge $-q_1$

The force on $-q_1$ is due to:

• the charge $+q_2$
• the charge $-q_3$

We need the $x$-component of the net force on $-q_1$.

Force due to $+q_2$ on $-q_1$

Since $-q_1$ and $+q_2$ are unlike charges, the force is attractive.

From the figure, this force acts along the horizontal line toward $+q_2$, so its entire contribution is along the $+x$ direction.

Its magnitude is proportional to $F_{12} \propto \frac{q_1 q_2}{b^2}$

Hence its $x$-component is proportional to $F_{12,x} \propto \frac{q_2}{b^2}$

(Here $q_1$ is common to all terms, so in the options it is omitted from proportionality.)

Force due to $-q_3$ on $-q_1$

Since $-q_1$ and $-q_3$ are like charges, the force is repulsive.

Thus the force on $-q_1$ is along the line joining $q_3$ and $q_1$, directed away from $q_3$.

If the line makes angle $\theta$ with the vertical/horizontal as shown, then from the geometry the horizontal component is proportional to $F_{13,x} \propto \frac{q_3}{a^2}\sin\theta$

and this component is toward the $+x$ direction.

Net $x$-component

Therefore, $F_x \propto \frac{q_2}{b^2} + \frac{q_3}{a^2}\sin\theta$

Matching with the options

This corresponds to: $\boxed{\frac{q_2}{b^2} + \frac{q_3}{a^2}\sin\theta}$

So the correct option is B.