Consider an embedding of $\overline{\mathbb Q}$ into $\mathbb C_p$ and $\alpha\in\overline{\mathbb Q}$ not rational.
Denote by $K/\mathbb Q$ the Galois extension generated by $\alpha$.
One can consider $K$ embedded in $\mathbb C_p$.
Let $\mathfrak p$ be a maximal ideal above $p$ of $O_K$, the ring of integers of $K$ and $x\in O_K$.
Denote by $v_p$ the valuation of $\mathbb C_p$, $v_{\mathfrak P}$ the valuation on $O_K$ associated to $\mathfrak p$ and $G$ the Galois group of $K/\mathbb Q$.
Do we have a relation between $v_{\mathfrak P}(x)$, $v_p(\sigma(x))$ ($\sigma \in G$) and $v_p(x)$?
Choose $\mathfrak P$ to be $\mathcal O_K \cap \mathfrak P_{\Bbb C_p}$, so the extension of $\Bbb Q_p$ generated by the image of $K$ is $K_\mathfrak P$, the completion of $K$ at $\mathfrak P$.
Then clearly $v_p(x) = v_{\mathfrak P}(x)$ for any $x \in K$.
Now for $\sigma \in G$, if $\sigma(\mathfrak P) = \mathfrak P$, then $v_p(\sigma(x)) = v_p(x)$; but otherwise one can find $x \in \mathfrak P \setminus \sigma(\mathfrak P)$, then $v_p(x) > 0$ while $v_p(\sigma(x)) = 0$, and vice versa, so nothing can be said about the relation.
Appendix
WLOG assume $\alpha$ is an algebraic integer and that $K = \Bbb Q(\alpha)$.
Let $f$ be the minimal polynomial of $\alpha$ over $\Bbb Q$. It is a monic polynomial in $\Bbb Z[x]$.
An embedding $K \to \Bbb C_p$ is determined by a root $\alpha'$ of $f$ in $\Bbb C_p$.
Factorize $f = f_1 \cdots f_r \in \Bbb Z_p[x]$ into monic irreducible polynomials, where we arrange the polynomials so that $f_1(\alpha') = 0$.
Let $F/\Bbb Q$ be the subfield generated by the coefficients of $f_1$, so $f_1$ is irreducible in $F[x]$. Note that $F \subset \Bbb Q_p$.
Then $[K:F] = [K_\mathfrak P : \Bbb Q_p] = \deg f_1$.
If $g_{\mathcal O_F \cap \mathfrak P}(K/F) > 1$ then we would have at least two $F$-embeddings $K \to \mathcal C_p$, so we would have $[K_\mathfrak P : \Bbb Q_p] < [K:F]$, which is a contradiction. Therefore, $g_{\mathcal O_F \cap \mathfrak P}(K/F) = 1$.
The properties $e_p(F/\Bbb Q) = f_p(F/\Bbb Q) = g_{\mathcal O_F \cap \mathfrak P}(K/F) = 1$ are enough for us to conclude that $F = K^{D_{\mathfrak P}}$, where $D_{\mathfrak P} \le G$ is the decomposition subgroup of $\mathfrak P$.
We summarize the behaviour of $p$ here: $$\begin{array}{rclcrcl} e_p(F/\Bbb Q) &=& 1 &~~~~~~~~~& e_p(K/F) &=& e_p(K_\mathfrak P / \Bbb Q_p) \\ f_p(F/\Bbb Q) &=& 1 && f_p(K/F) &=& f_p(K_\mathfrak P / \Bbb Q_p) \\ g_p(F/\Bbb Q) &=& [F:\Bbb Q] && g_p(K/F) &=& 1 \\ \end{array}$$
This provides an "algorithm": first factorize $(p)$ into product of distinct prime ideals in $F$, then choose one of them, say $\mathfrak p$. This gives rise to a unique prime ideal $\mathfrak P$ in $K$.
Then the extension $K_\mathfrak P / \Bbb Q_p$ can be "modelled" by $(\mathcal O_K)_{\mathfrak P} / (\mathcal O_F)_{\mathfrak p}$, i.e. by $K/F$ but we focus on the prime $\mathfrak p$, in the sense that the former is the completion of the latter.