It is AND gate implementation using NAND gate. To design $2$ input AND gate we required $2$ NAND gate.
Here the output of the first NAND gate is given as input for $2^{nd}$ NAND gate. We can write an equivalent expression for it.
$f=\overline{\left[\overline{(a.b)}.\overline{(a.b)}\right ]}$
$f=\overline{\overline{(a.b)}}+\overline{\overline{(a.b)}}$
$f=(a.b)+(a.b)$
$f=(a.b)$
$f$ is AND gate whose input is $a,b$
So Option $(C)$ is correct.