You are correct in saying that the splitting field can be written $\Bbb{Q}(a,i)$.
However there are three subgroups of order four.
1: The cyclic subgroup
The automorphism group contains a cycle of the roots, namely the map that takes $a$ to $\frac{i}{a}$ and $i$ to $-i$, which corresponds to the cycle $$\pmatrix{a & \frac{i}{a} & -a & \frac{-i}{a}}.$$
Now the subfield fixed by the group generated by this automorphism has degree 2 over $\Bbb{Q}$ by the main theorem of Galois theory. Note in addition that $ia^2+\frac{i}{a^2}$ is fixed by this map since $ia^2\mapsto -i(\frac{i^2}{a^2})=\frac{i}{a^2}$, and $\frac{i}{a^2}\mapsto -i(\frac{a^2}{i^2})=ia^2$.
However $$ia^2+\frac{i}{a^2}=i(1+\sqrt{2})+\frac{i}{1+\sqrt{2}}=i\left(1+\sqrt{2}+\frac{1-\sqrt{2}}{1-2}\right)=2i\sqrt{2},$$ which has degree two over $\Bbb{Q}$. Thus the fixed field corresponding to this subgroup is $$\Bbb{Q}\left(ia^2+\frac{i}{a^2}\right)=\Bbb{Q}[2i\sqrt{2}]=\Bbb{Q}[\sqrt{-2}].$$
2: A Klein 4 subgroup
Another subgroup of order four is generated by the automorphism that sends $a$ to $-a$ and $i$ to $i$ and the automorphism that sends $a$ to $a$ and $i$ to $-i$. Note that both of these fix $a^2$, which has order 2 over $\Bbb{Q}$, so the fixed field corresponding to this subgroup is $$\Bbb{Q}[a^2]=\Bbb{Q}[1+\sqrt{2}]=\Bbb{Q}[\sqrt{2}].$$
3: Another Klein 4 subgroup
Finally, there must be a third subgroup of order four, which is generated by the automorphism that sends $a$ to $\frac{i}{a}$ and $i$ to $i$ and the automorphism that sends $a$ to $-a$ and $i$ to $i$. Note that these both fix $i$, so the fixed field is $\Bbb{Q}[i]$.
Note that this makes a lot of sense, subgroups of order 4 correspond to fixed fields of degree 2 over the base field, or quadratic extensions. Therefore, we expect $\Bbb{Q}[i]$ and $\Bbb{Q}[\sqrt{2}]$, since both of these are clearly contained in the splitting field. Then we notice that since $\Bbb{Q}[\sqrt{2}]$ is a real field, and since $\Bbb{Q}[i]$ contains no real irrational elements, $\Bbb{Q}[i\sqrt{2}]=\Bbb{Q}[\sqrt{-2}]$ must be the last field (if you count subgroups of $D_4$, you can check there are 3 of order 4). Then we just have to match these fields up to groups.
Using the fundamental theorem of Galois theory here sounds like a good idea to me. Let $K = \mathbb{Q}(b, \zeta)$ in your notation. Since $P$ is a normal subgroup of $G$, $E := K^{P}$ is a Galois extension of $\mathbb{Q}$ with $\mathrm{Gal}(E/\mathbb{Q}) \cong G/P$. Moreover, $E$ is the unique extension of $\mathbb{Q}$ contained in $K$ satisfying $[E:\mathbb{Q}] = p-1$. Indeed, for any subfield $E' \subset K$ with $[E':\mathbb{Q}] = p-1$, then writing $E' = K^{H}$ for some subgroup $H$ of $G$, we see that $[G:H] = [E':\mathbb{Q}] = p-1$, which implies that $|H| = p$, i.e. $H$ is another Sylow $p$-subgroup of $G$. This forces $P = H$, since $P$ is the unique Sylow $p$-subgroup of $G$, and so $E' = E$.
Thus, it suffices to exhibit a degree $p-1$ extension of $\mathbb{Q}$ contained in $K$ which has cyclic Galois group over $\mathbb{Q}$, since this must be the equal to $E$ by the reasoning above. I leave to you to check that $\mathbb{Q}(\zeta)/\mathbb{Q}$ is one such extension.
Best Answer
$gfg^{-1}(2^{1/8})=gf(2^{1/8})=2^{1/8}g(e^{i\pi/4})=2^{1/8}e^{7\pi/4}=f^3(2^{1/8})$so $gfg=f^3$and G isn't D^8.