Given $: A=\left[\begin{array}{cc}\alpha & \beta \\ \gamma & -\alpha\end{array}\right]$

Then, $A^{2}=A A=\left[\begin{array}{ll}\alpha & \beta \\ \gamma & -\alpha\end{array}\right]\left[\begin{array}{ll}\alpha & \beta \\ \gamma & -\alpha\end{array}\right]$

$=\left[\begin{array}{ll}\alpha^{2}+\beta \gamma & \alpha \beta-\alpha \beta \\ \alpha \gamma-\alpha \gamma & \beta \gamma+\alpha^{2}\end{array}\right]$

$=\left[\begin{array}{cc}\alpha^{2}+\beta \gamma & 0 \\ 0 & \beta \gamma+\alpha^{2}\end{array}\right]$

Now, $A^{2}=1 \Rightarrow\left[\begin{array}{cc}\alpha^{2}+\gamma & 0 \\ 0 & \beta \gamma+\alpha^{2}\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$

On comparing the corresponding elements, we have:

$\alpha^{2}+\beta \gamma=1$

$\Rightarrow \alpha^{2}+\beta \gamma-1=0$

$\Rightarrow 1-\alpha^{2}-\beta \gamma=0$

Hence, the correct answer is $B$.