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.
Often, when doing circuit analysis, any current that enters one of the capacitor's plates is assumed to exit the other plate.
We can assume this because when we inject an electron on one plate, the field it produces will repel other free charges around it. If the nearest free charges are on the other plate, then those are the ones that will get repelled, leading to the current out of one terminal being equal to the current in the other.
Of course you can also arrange, for example, for both plates to have some potential relative to your reference ground node. If a net charge moves in or out of the capacitor to change this potential, then you would model that with a parasitic capacitance between the two terminals of your capacitor and some other location in the circuit. This parasitic capacitance would account for electric field lines that go from the capacitor structure to "somewhere else" rather than originating on one plate and terminating on the other.
one of my motivations for studying this is for high-frequency circuits
In high frequency circuits you won't be assuming that a metal object is an equipotential. If you make your two "plates" larger than ~1/10 of the wavelength associated with the highest frequencies in your circuit, you will create a distributed structure rather than a lumped one. If the "plates" are very long and skinny, you have made a transmission line, for example. Then you will find that signals propagate along the structure as waves, with behavior dictated by the balance of the capacitance and inductance of the structure.
At some level you should also remember that all of our lumped circuit analysis is an approximation, based on certain simplifying assumptions about the nature of the circuit. If the lumped circuit model of a capacitor isn't adequate for explaining some particular circuit or device, you may have to perform a more detailed analysis, for example using Poisson's equation to analyze an electrostatic structure, or Maxwell's equations to analyze situations where magnetic and electric fields interact with the structure of the circuit (i.e. high-frequency situations).
Best Answer
By Gauss's theorem, we have (Let $S$ be an area of a capacitor):
I will use your example in Edit. Let me call a plate with $2Q$ plate A, and the other plate B. Let a charge in downside of plate A be $q_1$, and upside of B $q_2$. Hence chrge on upside of A is $2Q - q_1$, and downside of B is $-Q-q_2$.
Now, consider the point inside of plate A. There is $2Q- q_1$ above, and $q_1 + q_2 + (-q_2 - Q) = q_1 - Q$ below. Thus, electric field inside plate A is $(3Q- 2q_1)/\epsilon_0 S$. Since this must be $E = 0$ (by induction!), $q_1 = 1.5Q$.
By doing the same in plate B, we have $q_2 = -1.5Q$.
Finally, consider an electric field between plates. Since, upside of A and downside of B (namely, outside of capacitor) have same charge $Q/2$, they don't create electric field. Charges inside of capacitor (downside of A and upside of B) create electric filed $E= 1.5Q/\epsilon S$. (Hence we have $1.5Q = C \times Ed$)
In general cases, let charge on A be $Q_1$, and charge on B be $Q_2$. Then, a charge upside of A and downside of B (outside of a capacitor!) are both $(Q_1 + Q_2)/2$, so they don't create an electric field inside the capacitor. We only need to consider charges inside the capacitor (downside of A and upside of B), which are $\pm(Q_1-Q_2)/2$. In this case, $$ \frac{Q_1 - Q_2}{2} = CV . $$