Let $K = \mathbb{Q}(\theta)$, where $\theta$ is a root of an irreducible polynomial $g \in \mathbb{Z}[t]$. Fix a rational prime $p$. Let $\theta^{(1)}, \ldots, \theta^{(n)}$ be the roots of $g$ in $\overline{\mathbb{Q}}$, and let $\phi^{(1)}, \ldots, \phi^{(n)}$ be the roots of $g$ in $\overline{\mathbb{Q}_p}$ (the algebraic closure of $\mathbb{Q}_p$). Let $f \in \mathbb{Z}[t]$, and put $$ \alpha = \frac{f(\theta^{(a)})}{f(\theta^{(b)})}, \qquad \beta = \frac{f(\phi^{(c)})}{f(\phi^{(d)})}. $$ Let $L$ be a finite extension of $K$ containing $\alpha$, and let $\mathfrak{p}$ be a prime ideal of $L$ above $p$.

If $\text{ord}_p(\beta) = 0$, is $\text{ord}_{\mathfrak{p}}(\alpha) = 0$ also? Why? What is the general relationship between $\text{ord}_p(\beta)$ and $\text{ord}_{\mathfrak{p}}(\alpha)$?

I know that if $\mathfrak{p}$ stands for a prime ideal of $K$ above $p$, if $g_{\mathfrak{p}}$ is the irreducible factor of $g$ in $\mathbb{Q}_p[t]$ which corresponds to $\mathfrak{p}$, and if $\phi^{(\cdot)}$ is any root of $g_{\mathfrak{p}}$, then for any $x \in K$ $$ \frac{1}{e}\text{ord}_{\mathfrak{p}}(x) = \text{ord}_p(\sigma(x)), $$ where $e$ is the ramification index of $\mathfrak{p}$ over $p$ and $\sigma$ is the embedding of $K$ into $\mathbb{Q}_p(\phi^{(\cdot)})$ that maps $\theta$ to $\phi^{(\cdot)}$ and fixes $\mathbb{Q}$ pointwise.

I have a feeling that I can view $\beta$ as the image of $\alpha$ under some embedding (there is an obvious candidate) and get a similar formula relating $\text{ord}_p(\beta)$ and $\text{ord}_{\mathfrak{p}}(\alpha)$, but I am having trouble proving this or making it concrete.

Can someone please help?