Let $p$ be an odd prime, and let $E/\mathbb{Q}_p$ be a finite extension that contains a primitive $p$-th root of unity $\zeta_p$ but not a primitve $p^2$-th root of unity $\zeta_{p^2}$. Let $a,b \in E^*$ such that $[E : \mathbb{Q}_p(a,b,\zeta_p)] = p^m$ for some $m \geq 1$.
Is the Hilbert symbol $(a,b)_p$ calculated over $E$ necessarily equals $1$?
No. Take $L/\mathbb{Q}_p$ the unique unramified extension of degree $p$.
Hence, the cyclic algebra $A'=(p,L/\mathbb{Q}_p,\sigma)$ is division (of degree $p$), where $\sigma$ is what you think. Since $\mathbb{Q}_p(\zeta_p)/\mathbb{Q}_p$ has degree $p-1$, which is prime to $p$, $A=A'\otimes_{\mathbb{Q}_p} \mathbb{Q}_p(\zeta_p)$ is still division of degree $p$. In fact, $A=(p,L(\zeta_p)/\mathbb{Q}(\zeta_p),\sigma')$, where $\sigma'$ is the canonical extension of $\sigma$.
The extension $L(\zeta_p)/\mathbb{Q}_(\zeta_p)$ is a cyclic Kummer extension of degree $p$, so you can write it down as $\mathbb{Q}_p(\zeta_p)(\sqrt[p]{d})/\mathbb{Q}_p(\zeta_p)$. Now take $a=p,b=d$, so that $\mathbb{Q}_p(a,b,\zeta_p)=\mathbb{Q}_p(\zeta_p)$.
By construction, $(a,b)_p$ is not $1$ over $\mathbb{Q}_p(\zeta_p).$
Now take $E/\mathbb{Q}_p(\zeta_p)$ any field extension of degree $p$ not containing $\zeta_{p^2}$ and which does not split $A$. Then $(a,b)_p$ is not $1$ over $E$.
