Step 1: Double inversion:
Two inversions in a row return the original signal.
Step 2: Apply:
A NOR is an OR followed by an inversion. Adding a NOT inverts again, so only the OR remains.
Step 3: Result:
$Y = A + B$, an OR gate, option (C).
Final Answer:
The circuit behaves as an OR gate.
\[ \boxed{\text{(C) }\text{OR},\ A+B} \]