This question is a digit-puzzle: we need two consecutive two digit numbers whose squares add up to a three digit number, with a very specific twist, the last digit of that sum has to match the tens digit of the smaller number, and none of the six letters P, Q, R, S, X, Y can repeat a digit.
Start from the size limit. If $a$ and $a+1$ are the two consecutive numbers, $a^2+(a+1)^2$ must stay under 1000 to remain a three digit number. Expanding, $2a^2+2a+1 \le 999$, so $a^2+a \le 499$. Testing $a=21$: $441+21=462$, fine. Testing $a=22$: $484+22=506$, too big. So only $a=10$ through $a=21$ can possibly work.
Now bring in the distinctness rule. Whenever $a$ and $a+1$ share the same tens digit, for example 14 and 15, the letters P (tens digit of $a$) and R (tens digit of $a+1$) come out equal, which the question forbids since all six letters must be different digits. Scan the list 10 to 21: every consecutive pair keeps the same tens digit, except the single pair 19 and 20, where the tens digit flips from 1 to 2. That makes $a=19$, $a+1=20$ the only pair worth checking.
With the full assignment confirmed, the letter Y equals 6, matching option (C).
\[ \boxed{Y = 6} \]