Measure the Phase Difference

The problem doesn't mention any LP filter or any way the average voltage is actually generated (as you noted).
It just asks what will be the average voltage (i.e. you don't have to assume any LP circuit; average voltage of a signal is defined even if there is no circuit actually generating the average voltage).
In other words: What will be the duty cycle of the output of the XOR gate? Multiply that by \$V\$ and you get the average voltage.

They are asking what is the average voltage (assuming 1 equals to V, and 0 to 0) as a function of the phase difference. You don't need to worry about the method they use for averaging.

Correct answer is A, because if the phase difference is 180deg, the output will be always 1 (=V) and the average will be V. If the phase difference is 90 deg, the output will be 0 half the time, and V half the time, thus average is 0.5V. If the phase difference is a bit smaller or greater than 180, the average voltage will be a bit smaller than V.

In my opinion the question is very poorly written in the paper.

Let the phase difference between \$f_a\$ and \$f_b\$ be \$\theta\$. Thus the average EX-OR over one cycle of \$f_a\$ would be

\$\theta*1+ (\pi-\theta)*0+\theta*1+ (\pi-\theta)*0 = 2\theta\$. (when \$ 0 <\theta<\pi\$) which tells that it should linearly increase as phase difference increase from \$0\$ to \$\pi\$.

\$(\theta-\pi)*0+ (2\pi-\theta)*1+(\theta-\pi)*0+ (2\pi-\theta)*1 = 2(2\pi-\theta)\$ (when \$ \pi <\theta<2\pi\$) which tells that it should linearly decrease as phase difference increase from \$\pi\$ to \$2\pi\$.

This corresponds to the graph in A.