Option C is correct here. Above circuit is Exclusive Nor Gate, Last Nor gate can be written as AND gate.
$f(x,y)= \overline{\overline{\left(((x+\overline{(x+y)})\right)}+\overline{\left(y+\overline{(x+y)})\right)}}$
$\qquad = ({x}+\overline{( x+ y)}). ({y}+\overline{(x+ y)})$
$\qquad =(x+\bar x\bar y).(y+ \bar x \bar y)$
$\qquad =(x+\bar y).(y +\bar x) = xy +\bar x \bar y$
$\qquad =x \odot y$
Refer here: Gate 1993
