What is the role of battery across a Parallel plate capacitor having equal charge? Each plate of a parallel plate capacitor has a charge $q$ on it. The capacitor is now connected to a battery. Now, which of the following options are correct?

a) The facing surfaces of the capacitor have equal and opposite charges.
  b) The two plates of the capacitor have equal and opposite charges.
  c) The battery supplies equal and opposite charges to the two plates.
  d) The outer surfaces of the plates have equal charges.

Before the battery is connected,
Applying Gauss Law, the inner surfaces of both the plates will have no charge on them, the outer surfaces will have charge $q$.
After the battery is connected,
The potential difference across the capacitor will gradually become equal to that of the battery .The inner surfaces will have equal and opposite charges according to $Q=CV$ where $V$ is the potential difference across the battery.
But, what will be the charges on the outer surfaces of the two plates ?
I am not clear about the role of battery in affecting the charges on the surfaces of the plates.
Option a) Looks right - Not sure about the other options.
Kindly explain what happens to the plates of the capacitor after the battery is connected ?
Does a battery always produce equal amount of negative and positive charges or it may produce additional charges ?
 A: 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.
