Suppose that $a,b \in \overline{\mathbb{Q}}$ are such that the field extension $\mathbb{Q}(b) \mid \mathbb{Q}(a)$ is normal. Further, let $\mathfrak{p} \neq (0)$ be a prime ideal in the ring of integers $\mathcal{O}_{\mathbb{Q}(a)}$ of $\mathbb{Q}(a)$ so that the minimal polynomial of $b$ over $\mathbb{Q}(a)$ splits over the completion $\mathbb{Q}(a)_{\mathfrak{p}}$. If $P$ is a prime ideal in $\mathcal{O}_{\mathbb{Q}(b)}$ such that $P \cap \mathcal{O}_{\mathbb{Q}(a)} = \mathfrak{p}$, is it then correct that the completions agree, i.e. that $$\mathbb{Q}(a)_{\mathfrak{p}} \cong \mathbb{Q}(b)_P?$$
EDIT: If this simplifies the matter, we may stick to the case $a=0$, thus $\mathbb{Q}(a) = \mathbb{Q}$.
Yes, this is correct.
More generally, let $ L/K $ be an algebraic extension where $ L=K(\alpha) $ and $ \alpha $ has minimal polynomial $ f(X) $ over $ K $. Let $ v $ be a valuation on $ K $. Then, the valuations on $ L $ which extend $ v $ are in one-to-one correspondence with the irreducible factors of $ f(X) $ factored in $ K_v[X] $. That is, if $ f(X) = \prod_{i=1}^{r} f_{i}(X)^{m_i} $ is the factorization of $ f $ into irreducibles in $ K_v[X] $, there are exactly $ r $ extensions of $ v $ to $ L $, say $ w_1, w_2, \ldots ,w_r $ and for each completion, we have an isomorphism $$ L_{w_i} \cong \frac{K_v[X]}{f_i(X)} \ \ , \ \ \ \forall \ i = 1,2 \ldots,r$$ In the case in question, the minimal polynomial splits, so the $ f_i $ are linear factors and hence you get the desired isomorphism. Details of this can be found in Chapter 2 of Neukirch's book, Algebraic Number Theory.