Since the answer has already been spoon fed to you (although edited out now, so I'll take out my references too) I guess I can now give you the reasoning behind the answer, and also expand on my comment to the question.
Of the four possible answers, all four are present in the circuit in different places at different times.
The circuit can be seen, in this situation, as being in two possible states:
simulate this circuit – Schematic created using CircuitLab
In the left hand circuit the diode is reverse biased, and no current will flow through it. In the right hand circuit it is forward biased and current will flow through it. Let's take the right-hand, forward biased, circuit first:
Current flows through the diode, so there will be a voltage drop of -0.7V across it. It's -0.7V since ground is the +5V point of the battery, so all the voltages are below that, i.e., negative WRT ground. With -0.7V across the battery that leaves -4.3V left to be across the resistor R. So that is your answers D and B accounted for respectively.
So let's look at the left hand circuit looking at the same values.
The diode is reverse biased. No current will flow through it. Therefore no current will flow through the resistor either. Since V=R×I, and I is 0, then V must also be 0, regardless of the resistance. So the voltage across the resistor is 0V (answer C), which leaves the battery's full 5V left across the diode. That's answer A.
But which is the right answer for your question? Well, as the others have noted, the question is badly worded. What it should be asking is not what is the Peak Inverse Voltage of the diode, but the Peak Inverse Voltage that gets applied to the diode.
The Peak Inverse Voltage is the maximum it can cope with when reverse biased. So that means that it must be the left hand, reverse biased, circuit we are interested in. The voltage applied across the diode in that circuit, as we have already discovered, is 5V.
So which is the right answer to your question do you think?