Please help, why is the answer D? This is similar to the circuit question I asked before?

Just treat the voltmeter as a "resistor of very high resistance" and an ammeter as a "resistor of very low resistance". Then the circuit can be regarded as a 2-volt battery connected to a 100-ohm resistance, and then in series with a "high resistance resistor" (represents the voltmeter) and a "low resistance resistor"(represents the ammeter) in parallel.

(a) When switch K is opened, the circuit becomes a 2-volt battery connected to a 100-ohm resistance and a "high resistance resistor" in series. Clearly, the "high resistance resistor" will share most the the voltage from the 2-volt battery. Because a voltmeter records the p.d. across its own internal resistance, the voltmeter reading is close to 2 volts.

(b) When switch K is closed, the voltmeter and ammeter are in parallel. Remember the fact that when resistors are in parallel, the equivalent resistance must be lower than the smallest resistance in the parallel circuit.

Hence, the equivalent resistance of the voltmeter-ammeter parallel arrangement must be smaller than the resistance of the ammeter. Since the resistance of the ammeter is already very small, the equivalent resistance of the parallel arrangement should be much lower than 100-ohm. As such, most of the voltage drop from the 2-volt batter will be taken by the 100-ohm resistor, leaving only a very small potential drop at the parallel arrangement. As said before, the voltmeter records the potential drop across its own internal resistance, the reading shown by the voltmeter is thus close to 0 volt.

Note that in the voltmeter-ammeter parallel arrangement, the current flows through the ammeter is larger than that through the voltmeter (due to the small resistance of the ammeter), but the p.d. across both the ammeter and voltmeter are the same. (just compare it with the question you asked before).

In the question that you asked before, the voltmeter is more or less directly connected to the two terminals of the battery (if we take the resistance of the light bulb to be small compared with the resistance of the voltmeter, and its presence could somehow be neglected). The voltmeter thus measures the voltage of the battery.

But in this problem, one terminal of the voltmeter is NOT connected to the battery. Hence, its reading wouldn't be the battery voltage.

Suppose the terminal of the voltmeter that is connected to the right-hand side of the 100-ohm resistor is now re-connected to the left-hand side of that resistor, then the voltmeter will show a reading of 2 volts (the battery voltage) whether switch K is opened or closed.

In the ideal condition. The voltmeter's resistance =Unlimited, The ammeter's resistance =0
so when K is closed ,it's short circuit(voltmeter=0v, 200Ohm=2v) then K is open, it liked open circuit(voltmeter=2v, 200Ohm=0v)