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.
For a capacitor with closer plate spacing (all else being equal), the electric field in the dielectric between the plates is stronger.
Assuming a uniform electric field, the potential difference is given by the product of the spacing and the strength of the electric field:
$$\Delta V = E\cdot d $$
So, the potential difference can be the same for both capacitors even though the plate spacing is different but the electric field between the plates will be different.
This is due to the fact that, for smaller spacing, the capacitance $C$ is larger and, thus, more charge $Q$ must be moved by the battery to charge the capacitor to 9V.
$$\Delta V = \frac{Q}{C}$$
Best Answer
This is based on a false premise. There is no rule that says that "potential differences are equal across two capacitors in series".
In a lumped circuit model, any two devices in parallel must have the same potential across them. This is because of Kirchoff's voltage law (KVL) which says that the net potential drop around any loop in a planar circuit is zero. To put it in more basic terms, if the potential at point 'A' is $V_a$ and the potential at point 'B' is $V_b$, then the potential difference between points A and B is $V_a - V_b$ no matter what path you take through the circuit between those points.
As arvindpujari's answer points out, since the potential differences are equal across the two parallel capacitors, this means that if the capacitance values aren't equal then they must have different charges since a linear capacitor is defined by the equation $V=Q/C$.
Imagine you start with two capacitors in series with no charge ($V=0$). Then start driving a current through them. The same current will flow through both capacitors (that's what it means for two elements to be in series), so the same charge will accumulate on each capacitor's plates.
Since they have equal charge, if the capacitance values are not the same, then the potentials must also not be the same.