Right answer is (b) e^{-t} \{cos(2\sqrt{2t})-sin(2\sqrt{2t})\}
The best I can explain: In the given question,
L^{-1} \left (\frac{s}{2s+9+s^2}\right )=L^{-1} \left (\frac{s}{(s+1)^2}+8)\right )
=L^{-1} \left (\frac{(s+1)-1}{(s+1)^2+8}\right )
=e^{-t} L^{-1} \left (\frac{(s-1)}{s^2+8}\right ) ———————–By First Shifting Property
=e^{-t} L^{-1} \left (\frac{s}{s^2+8}\right )-e^{-t} L^{-1} \left (\frac{1}{s^2+8}\right )
=e^{-t} \{cos(2\sqrt{2t})-sin(2\sqrt{2t})\}.