So I showed that $\mathbb{Q}_{p}[\theta]$ is an unramified extension of degree p, where $0=g(\theta)=\theta^{p}-\theta-1$.
But it also follows that $\mathbb{Q}_{p}[\phi]$ is an unramified extension of degree p, where $0=f(\phi)=\phi^{p}-\phi-a$ and $|a|_{p}=1$ .
So doesn't just follow that $\mathbb{Q}_{p}[\theta]=\mathbb{Q}_{p}[\phi]$ from the 1-1 correspondence of unramified extensions of a local field? Thanks
For every local field $K$ and natural number $n$ coprime to $K$'s residue characteristic, there is a unique unramified extension $L/K$ of degree $n$. (Have you proven this for $K=\Bbb Q_p$?)
Since $\Bbb Q_p[\theta]$ and $\Bbb Q_p[\phi]$ are both unramified extensions of degree $p$ they are equal, yes.