Ignore inner and outer surfaces. There is just one surface.
Imagine a single, infinite plane with some positive charge density. You can easily show there would be an electric field of constant strength*, perpendicularly out of the plane all the way to infinity on both directions.
Now imagine a single, infinite plate with the same negative charge density. There would be an electric field of constant strength perpendicularly into the plane all the way to infinity in both directions.
Put these two plates on top of each other, and these fields perfectly cancel.
Put these two plates in parallel, and because the field is constant strength it will perfectly cancel everywhere except between the two plates, where the electric field directions are the same and it will add to be twice as strong.
[*By constant strength I mean the electric field is just as strong no matter how far you are from the plate. Why is the field constant strength? Because the field lines can't ever diverge from one another. The way fields usually get weaker is the equipotential surface the field lines are normal to gets bigger as you increase the distance from the object. So the same number of field lines piercing a bigger surface means a field lines are more spread out, and thus a weaker field. In this case however, the equipotential surfaces are always a pair of infinite parallel planes, no matter what distance we are from the charged plane. No spreading means no change in field strength.]
I think you question can be answered succinctly from this point:
Does a battery always produce equal amount of negative and positive charges or it may produce additional charges ?
A battery never produces additional charge. Very few things can do this (like a Van de Graaff generator), and even those can only do so locally. A battery pumps charge, and that leaves no room for adding to the total charge count (c).
I thought it strange that the problem is starting out saying that the capacitor has on net extra charge, but you have analyzed the situation correctly. Since we're treating it as an infinite parallel plate capacitor, before the battery is connected the charge is concentrated on the outside face of both plates.
When the battery is connected, and the capacitor has fully charged (if you neglect internal resistance this last specifier isn't needed), then you will have additional positive charge on one plate and additional negative charge on the other. Following the eqi-potential principle, these opposite charges will gather on the inner faces of the two plates (a).
Now a deeper question: will connecting the battery affect the charge density on the outer faces of the plates? I will argue "no". The important unwavering assumption is a constant electric potential throughout the interior of a plate. Introducing the battery introduces a field in-between the inner surfaces, which results in potential difference equal to the voltage of the battery. You have 4 points (which are infinite planes) along the number line (since this is 1D symmetry) where charge is located, and the field is $E=\sigma/(2 \epsilon_0)$ pointing away from the plane - that is, proportional to the charge density and constant. That means that the field on the outside of the plates won't change when connecting the battery, since the total charge quantity stays the same, the field doesn't diminish with distance from the charges, and you only moved charges around. The field within a plate is zero and the outside charge density remains the same, thus, in order to transition from a zero field to the unchanged outside field strength, you require the same charge density on the outside face (d). The potential on the surface is different, but the charge is the same - a very important nuance of electronics.
Thus, options "a", "c", and "d" are both correct.
Best Answer
First, to be clear, when we speak of a charged capacitor, we almost never mean electrically charged but, rather, 'charged' with energy in the same way we speak of a charged battery.
If the plates of a capacitor with capacitance $C$ have equal and opposite electric charge $Q$, the capacitor is electrically neutral but stores an energy
$$U = \frac{Q^2}{2C} = \frac{1}{2}QV$$
where $V$ is the potential difference between the plates.
If the plates of a capacitor have unequal charge, there is now energy stored in more than one capacitance. There is the capacitance that exists between the two plates (the mutual capacitance) as well as the capacitance of each plate to the environment.
Let $V_1$ be the potential of plate 1 with charge $Q_1 = Q$.
Let $V_2$ be the potential of plate 2 with charge $Q_2 = -Q + \delta Q$.
The energy of system is then
$$U = \frac{1}{2}Q_1V_1 + \frac{1}{2}Q_2V_2 = \frac{1}{2}Q(V_1 - V_2) + \frac{1}{2}\delta QV_2 = \frac{Q^2}{2C} + \frac{\delta Q^2}{2C_{20}}$$
where $C_{20}$ is the capacitance of plate 2 to the environment.