Truth table for the system on four NAND gates as shown in figure is

 

$\gamma =\overline{\left(\overline{\left(\overline{A\cdot B}\right)\cdot A}\right)\cdot \left(\overline{\left(\overline{A\cdot B}\right)\cdot B}\right)}$

Let A=0 & B=0                                           Let A = 1 & B = 0

$$\begin{gathered} \gamma =\overline{\left(\overline{\left(\overline{A\cdot B}\right)\cdot A}\right)\cdot \left(\overline{\left(\overline{A\cdot B}\right)\cdot B}\right)} \gamma =\overline{\left(\overline{\left(\overline{A\cdot B}\right)\cdot A}\right)\cdot \left(\overline{\left(\overline{A\cdot B}\right)\cdot B}\right)} \\ =\overline{\left(\overline{\left(\overline{0.0}\right)\cdot 0}\right).\left(\overline{\left(\overline{0.0}\right)\cdot 0}\right)} =\overline{\left(\overline{\left(\overline{1.0}\right)\cdot 1}\right)\left(\overline{\left(\overline{1.0}\right)\cdot 0}\right)} \\ =\overline{\left(\overline{\overline{0}.0}\right).\left(\overline{\overline{0}.0}\right)} =\overline{\left(\overline{\overline{0}.1}\right).\left(\overline{\overline{0}.0}\right)} \\ =\overline{\left(\overline{1.0}\right).\left(\overline{1.0}\right)} =\overline{\left(\overline{1.1}\right).\left(\overline{1.0}\right)} \\ =\overline{\left(\overline{0}\right)\cdot \left(\overline{0}\right)} =\overline{\left(\overline{1}\right).\left(\overline{0}\right)} \\ =\overline{1.1} =\overline{\left(0\right).\left(1\right)} \\ =\overline{1} =\overline{0} \\ = 0 = 1 \end{gathered}$$

Let A=0 & B=1                                           Let A = 1 & B = 0

$$\begin{gathered} \gamma =\overline{\left(\overline{\left(\overline{A\cdot B}\right)\cdot A}\right)\cdot \left(\overline{\left(\overline{A\cdot B}\right)\cdot B}\right)} \gamma =\overline{\left(\overline{\left(\overline{A\cdot B}\right)\cdot A}\right)\cdot \left(\overline{\left(\overline{A\cdot B}\right)\cdot B}\right)} \\ =\overline{\left(\overline{\overline{0.1}.0}\right).\overline{\left(\overline{0\cdot 1}\right)\cdot 1})} =\overline{\left(\overline{\overline{1.1}.1}\right).\overline{\left(\overline{1\cdot 1}\right)\cdot 1})} \\ =\overline{\left(\overline{\overline{0}.0}\right).(\overline{\overline{0}.1)}} =\overline{\left(\overline{\overline{1}.1}\right).(\overline{\overline{1}.1)}} \\ =\overline{\left(\overline{1.0}\right).\left(\overline{1.1}\right)} =\overline{\left(\overline{0.1}\right).\left(\overline{0.1}\right)} \\ =\overline{\left(\overline{0}\right).\left(\overline{1}\right)} =\overline{\left(\overline{0}\right).\left(\overline{0}\right)} \\ =\overline{\left(1\right).\left(0\right)} =\overline{1.1} \\ =\overline{0} =\overline{1} \\ =1 =0 \\ \begin{array}{lll}A& B& t\\ 0& 0& 0\\ 0& 1& 1 option 1 is correct\\ 1& 0& 1\\ 1& 1& 0\end{array} \end{gathered}$$