The net charge of any of those internally connected pairs of plates is always zero. That is, when you charge the capacitors, charge doesn't leave the wire between C and D, it only moves along it, and is held in place by the electric field of the adjacent plates. If a circuit is completed that allows charge to flow from D's negative plate to A's positive plate, the charges will move back to the right place, but the net charge of the 4 capacitors will always be the same.
Connecting the positive terminal of A will not allow charge to flow back from D, so nothing will happen. Similarly, connecting the wire between C and D won't make charge flow in or out of it, at least not in any way significant to the circuit. It only changes the reference for where we make our measurements from.
how can we calculate the terminal voltage?
As the problem is stated, the question answer itself:
the capacitor is connected to a power source with constant voltage
$V_s$
(emphasis mine). By KVL, the voltage across the capacitor is the voltage across the power source which is the constant $V_s$.
Suppose we move the plates closer to distance d2. Will the terminal
voltage be identical to that of d1?
Yes, since the voltage across the capacitor is the constant $V_s$.
However, since changing the distance between the plates changes the capacitance, there will be a current through the capacitor proportional to the rate of change of capacitance:
$Q = CV$
$\dfrac{dQ}{dt}= I_c = C\dfrac{dV_s}{dt} + V_s \dfrac{dC}{dt}$
Since the voltage is stipulated to be constant, the first term on the right hand side is zero.
However, even with a constant voltage across the capacitor, there can be a capacitor current due to time dependent capacitance.
Best Answer
Every system likes to decrease its electrostatic energy.
The charges on the plates are almost in stable equilibrium. The charges on the opposite plates attract them, and the charges on the same plate repel them with almost the same force.
However, a capacitor has fringe fields:
These may be negligible when calculating the field inside a capacitor, but they are extremely important when there are wires in play -- by $\vec J=\sigma\vec E$, for a wire (which has high $\sigma$), even a small $\vec E$ can create a large current.
And these create the tiny perturbation required to push the charges out of their almost-stable equilibrium.
Note that redistribution of surface charges occurs on the wires as well, so it's not just the fringe fields pushing the electrons.