Here, $\omega^5=1\Rightarrow \omega^4=\omega^{-1}\Rightarrow \omega^3=\omega^{-2}\Rightarrow \omega^2=\omega^{-3}$
$\therefore \omega^{5-k}=\omega^{-k}$ and $\omega^{5n}=1,~ n \in \mathbb{Z}$.
Now
$\begin{align} \sum_{k=0}^{4}b_k \omega^{k} &= \sum_{k=0}^{4}\left( \sum_{j=0}^{4}j \omega^{-kj} \right) \omega^{k} \\ &=\sum_{k=0}^{4}\sum_{j=0}^{4}j \omega^{k-kj}\\ &=\sum_{k=0}^{4} \left(0+ \omega^{0}+2 \omega^{-k}+ 3\omega^{-2k}+ 4\omega^{-3k} \right)\\ &=\sum_{k=0}^{4} \left(1+2 \omega^{4k}+ 3\omega^{3k}+ 4\omega^{2k} \right)\\ &=\sum_{k=0}^{4}1+2\sum_{k=0}^{4}(\omega^{4})^{k}+3\sum_{k=0}^{4}(\omega^{3})^{k}+4\sum_{k=0}^{4}(\omega^{2})^{k}\\ &=5+2\frac{(\omega^4)^5-1}{\omega^4-1}+3\frac{(\omega^3)^5-1}{\omega^3-1}+4\frac{(\omega^2)^5-1}{\omega^2-1}\\ &=5+2\frac{\omega^{20}-1}{\omega^4-1}+3\frac{\omega^{15}-1}{\omega^3-1}+4\frac{\omega^{10}-1}{\omega^2-1}\\ &=5+0+0+0; ~[\because \omega^{5n}=1]\\ &=5 \end{align}$
So the correct answer is A.