$S$ = {necklaces of length 14 with 8 blue, 3 green and 3 brown beads}.
Clearly,$|S| = \frac {14!} {8!3!3!} $.
The group of symmetries, $G = D_{14}$. Clearly, $|G| = 28$. And as you have identified G is acting on S.
And by Burnside lemma, required answer is, $$ \#orbits = \frac 1 {|G|}*\sum_{\sigma \in G}fix(\sigma)$$
1) $\sigma = identity $
Then it fixes any element in $S$. Thus, $fix(\sigma) = |S| = \frac {14!} {8!3!3!}$
2) Rotations, $\sigma$ = rotation by $\frac {360} {14} degree$ clockwise = $p^1$
Carefully consider the cyclic structure of the permutation $\sigma$,
$$\begin{pmatrix}
1&2&3&4&5&6&7&8&9&10&11&12&13&14\\
2&3&4&5&6&7&8&9&10&11&12&13&14&1
\end{pmatrix}$$
$$\equiv (1\ 2\ 3\ 4\ 5\ 6\ 7\ 8\ 9\ 10\ 11\ 12\ 13\ 14) \ \ \text{[in cycle notation]}$$
If $\sigma$ fixes $x \in S$, then all vertex of the 14-gon must have the same color which is not the case, hence $fix(\sigma)=0$.
Clearly, $fix(\sigma=p^{13})=0$.
3)$\sigma=p^2$ i.e
$$\begin{pmatrix}
1&2&3&4&5&6&7&8&9&10&11&12&13&14\\
3&4&5&6&7&8&9&10&11&12&13&14&1&2
\end{pmatrix}$$
$$\equiv (1\ 3\ 5\ 7\ 9\ 11\ 13)(2\ 4\ 6\ 8\ 10\ 12\ 14) \ \ \text{[in cycle notation]}$$.
So, $\sigma$ has two cycles of length 7 and if you think carefully, for x to be in $fix(\sigma)$ all vertices in a single cycle will have same bead color. Which is not possible with 8 red, 3 blue and 3 brown beads. So, $fix(\sigma)=0$.
Clearly, $fix(\sigma=p^{12})=0$.
In the same fashion compute the $fix(\sigma)$ for the remaining by looking into the cycle structure and then use the burnside to get the required answer.