Let f(x)=

$f(x)=\ln (\sin x), g(x)=\sin ^{-1}\left(e^{-x}\right)$

$\Rightarrow f(g(x))=\ln \left(\sin \left(\sin ^{-1} e^{-x}\right)\right)=-x$

$\Rightarrow f(g(x))=-\alpha$

But given that $(f o g)(\alpha)=b$

$\therefore-\alpha=b$ and $f^{\prime}(g(\alpha))=a$, i.e., $a=-1$

$\therefore a \alpha^{2}-b \alpha-a=-\alpha^{2}+\alpha^{2}-(-1)$

$\Rightarrow a \alpha^{2}-b \alpha-a=1$