Let, the roots of equation
$2 x^3+a x^2-8 x+b=0$ are $\alpha, \beta, \gamma$
After reducing by one from the roots, the quantities becomes $\alpha-1, \beta-1, \gamma-1$.
Now, let
$x=\alpha-1$
$\Rightarrow \quad \alpha=x+1$
Since, $\alpha$ is the root of given equation, so
$\begin{aligned} & 2(x+1)^3+a(x+1)^2-8(x+1)+b=0 \\ & \Rightarrow 2\left(x^3+3 x^2+3 x+1\right)+a\left(x^2+2 x+1\right) \\ & -8(x+1)+b=0 \\ & \Rightarrow 2 x^3+(6+a) x^2+(6+2 a-8) x\end{aligned}$
$+(2+a-8+b)=0$
According to given informations
$6+a=0 \Rightarrow a=-6$
and $\quad 2+a-8+b=0 \Rightarrow b=12$
So
the given equation becomes
$2 x^3-6 x^2-8 x+12=0$, by trial and error method $(x-1)$ is the factor of the equation, so
$\begin{aligned}\left(2 x^3-6 x^2-8 x+12\right)=(x-1) & \left(2 x^2-4 x-12\right) \\ = & 2(x-1)\left(x^2-2 x-6\right)\end{aligned}$
So,
roots of the equation $2 x^3-6 x^2-8 x+12=0$,
are $1, \frac{2 \pm \sqrt{4+24}}{2}=1,1 \pm \sqrt{7}$ Hence, option (b) is correct.