a
(a) \(C = \overline {\bar A.\bar B} = \overline {\bar A + \bar B} = A + B\) (De morgan’s theorem)
Hence output \(C\) is equivalent to \(OR\) gate
\(C = \overline {\overline {AB} .\overline {AB} } = \overline {\overline {AB} + \overline {AB} } = AB + AB = AB\)
In this case output \(C\)  is equivalent to \(AND\)  gate.