Is there an isomorphism between $\text{Spec}(R_{\mathfrak{p}})$ and the prime ideals of $R$ which are contained in $\mathfrak{p}$?

Suppose $$R$$ is a ring, and $$\mathfrak{p} \in \text{Spec}(R).$$

I have been told that $$\text{Spec}(R_{\mathfrak{p}}) \cong \lbrace \mathfrak{q} \in \text{Spec}(R)\;| \mathfrak{q} \subset \mathfrak{p} \rbrace,$$ and I am trying to figure out whether this is true or not, but I am stuck trying to show that there is an injective map $$\alpha : \text{Spec}(R_{\mathfrak{p}}) \longrightarrow \lbrace \mathfrak{q} \in \text{Spec}(R)\;| \mathfrak{q} \subset \mathfrak{p} \rbrace.$$

I suspect that there is a way to show this by exploiting the fact that any (prime) ideal of $$\text{Spec}(R_{p})$$ cannot contain units of $$R_{\mathfrak{p}}$$, but I can't find a way through.

Any assistance would be much appreciated.

In general, for a multiplicatively closed subset $$S$$ of $$R$$, there is a natural correspondence between the prime ideals of $$S^{-1}R$$ and the prime ideals of $$R$$ disjoint from $$S$$. Your example is the case where $$S=R-\newcommand{\fp}{\mathfrak{p}} \newcommand{\fq}{\mathfrak{q}}\fp$$.
The prime ideals of $$S^{-1}R$$ all have the form $$S^{-1}\fq$$ where $$\fq\subseteq \fp$$ is a prime ideal of $$S$$. It is straightforward to check that these $$S^{-1}\fq$$ are prime in $$S^{-1}R$$. Conversely, let $$Q$$ be a prime ideal of $$S^{-1}R$$. Then $$\fq=\{a\in R:a/1\in Q\}$$ is a prime ideal of $$R$$, being the inverse image of $$Q$$ under the map $$\phi:a\mapsto a/1$$ from $$R$$ to $$S^{-1}R$$. If $$b/s\in Q$$, then $$b/1=(b/s)(s/1)\in Q$$ so $$b\in\fq$$ and so $$Q=S^{-1}\fq$$ etc.
Your $$\alpha$$ is $$Q\mapsto\phi^{-1}(Q)$$.