Behavior of this RC circuit

Question: once the capacitors are charged, the switch is moved to position Y. State what happens to the p.d across Capacitor P and Q

The instant after switching occurs, assuming ideal capacitors, the p.d. across capacitors P and Q will be the same as the p.d. across these capacitors the instant before switching occurs. The reasons are as follows:

The relationship between capacitor voltage, current and capacitance is given by

$$V=frac{Q}{C}$$

So for a given capacitance $C$, the voltage across the capacitor is proportional to the net charge $Q$ on each plate of the capacitor (which are equal in magnitude and opposite in polarity). This means charge must be moved from one plate to the other to change the net charge on each plate and the voltage. That requires current and time according to

$$v(t)=frac{1}{C}int i(t)dt+V_i$$

TO some extent, I understand that the capacitor P, with the help of the new path for the charges, will discharge itself. But I really dont get why all of the p.d must be across Q, and none of it across the resistor.

Under steady state conditions in a dc circuit, ideal capacitors look like open circuits and no current can flow to the capacitor. The relationship between capacitor current and voltage is

$$i(t)=Cfrac{dv(t)}{dt}$$

Under steady state conditions voltages in a dc circuit are not changing in time. Therefore $dv(t)/dt=0$ and $i(t)=0$. Therefore when the switch is in position Y for a long time, there is no current flowing in the series circuit of capacitor Q and resistor R. If there is no current flowing through resistor R then the p.d. across resistor R is zero. Since capacitor P is in parallel with resistor R, there is no p.d. across capacitor P. That, in turn, means all the voltage of the two batteries is across capacitor Q.

Also, other than the explanation of the mechanism using laws of electrostatics, is there any way to predict the p.d s in the steady state directly using circuit equations? I really don't know how resistors and capacitors work together in RC circuits

You can predict the voltage and current for an ideal capacitor using the last two equations I gave you. When you analyzed dc circuits with capacitors (and/or inductors) you need to separate the analysis into different time periods, using these steps:

1. Steady state conditions prior to a switching event. Under these conditions all ideal capacitors look like open circuits because voltages are not changing and current to a capacitor is zero per the third equation above.

2. Switching event time t=0. The instant after switching the voltages across all capacitors are the same as they were the instant before switching. The reason is you can't change the voltage across an ideal capacitor in zero time, per the second equation above. It takes time and current to change the charge Q, and thus the voltage V in the first equation above.

3. The transient period t=0 to t=∞ (steady state). During this period voltages and currents are changing as capacitors charge and/or discharge. You can look up the equations covering these changes for dc RC circuits.

4. Resumption of steady state conditions (t=∞). Reapply step 1 with the switch in the new position.

Hope these help you.