Does an accelerating airplane also start climbing?
You assume that three of the four forces are constant. This is not quite right: Only the mass is given, all the other forces are variable.


Thrust $T$: Changes with throttle setting. Also, thrust can vary with flight speed $v$ in a way that can be expressed as $T \propto v^{n_v}$.
Lift $L$: Depends on speed and on angle of attack $\alpha$, as in $L \propto v^2 \cdot (\alpha - \alpha_0)$. The accelerating airplane is held level by gradually reducing the angle of attack, such that lift exactly equals weight. The elevator is the instrument for controlling angle of attack.
Drag $D$: Like lift, drag depends also on speed and angle of attack: $D \propto v^2\cdot (\alpha - \alpha_0)^2$


Your option a is the correct one: The pilot compensates by adjusting the angle of attack by gradually pushing the stick forward. If he would keep the stick (and consequently the elevator) unchanged, the airplane would climb.

Now it might be opportune to introduce the concept of specific excess power ($SEP$). This is the extra power available to the pilot, related to the aircraft weight to make it comparable between airplanes. $$SEP = \frac{P_{eff}-D\cdot v}{m\cdot g}$$
where $P_{eff}$ is the effective power: $P_{eff} = T\cdot v$. When the pilot commands more thrust than needed for level flight, the excess thrust can be either converted into kinetic energy (speed) or potential energy (altitude). If you look at the units, $SEP$ is a speed. It tuns out that it is the possible climb speed available if flight speed stays constant: $$SEP = v_z$$
In reality, the practical climb speed is smaller because the airplane needs to speed up when climbing to compensate for the decreasing air density.

The possible acceleration is almost as easy to calculate from the specific excess power: $$\ddot x = SEP\cdot\frac{g}{v}$$

Of course, a mixture of both is also possible, as is trading in one for the other. By adjusting the elevator and with it the flight path, the pilot decides what happens.