In Digital, We are not interested in exact values of input and output values. We just consider $1$ as high and $0$ as low.
Positive logic system : In positive logic system $1$ is considered as High and $0$ is considered as Low.
Negative logic system : In negative logic system $0$ is considered as High and $1$ as Low.
So, a circuit which behaves as $NAND$ gate in $+ve$ logic system is $F(A,B) = \overline{AB}$, And when this is considered in $-ve$ logic system replace $0$ by $1$ and $1$ by $0$. i.e., replace a complemented variable by uncomplemented and uncomplemented by complemented.
$F = \overline{AB}$
Replace $A\rightarrow \overline{A}$ , $B\rightarrow \overline{B}$ and $\overline{AB}\rightarrow AB $ and see how circut behaves.
$F = \overline{\overline{\overline{A}.\overline{B}}}$ $ = \overline{A+B}$, So thus the circuit behaves as $NOR$ gate.
Hence (B) is correct answer.
This question can also be done by forming the truth table for NAND logic first and then Swapping $0$ and $1$ and Analyise how circuit behaves. ( but it takes some time )
