If $R$ is a PID and $P$ is a nonzero prime ideal, then $E(R/P)=K/P_P$, where $K$ is the fraction field of $R$ and $R/P$ is viewed as a submodule of $K/P_P$ via the evident map. Indeed, one readily checks that $R/P\cong R_P/P_P$ is an essential submodule of $K/P_P$ and that $K/P_P$ is divisible, hence injective (injective and divisible are the same since $R$ is a PID).

This fact can be generalized to arbitrary ideals in Dedekind domains:

**Theorem.** Let $R$ be a Dedekind domain with fraction field $K$ and let $I$ be an ideal of $R$ different from $0$ and $R$. Let $P_1,\dots,P_t$ denote the primes containing $I$ and let $S$ be the multiplicative set $R\setminus (P_1\cup\dots\cup P_t)$. Then the natural map $R/I\to K/S^{-1}I$ is an injective envelope of $R/I$.

*Proof.* Observe that the image of any $s\in S$ in $R/I$ is invertible. Indeed, it is not contained in any maximal ideal of $R/I$. Thus, $R/I$ is $S$-divisble, and it follows that the natural map $R/I\to S^{-1}(R/I)=S^{-1}R/S^{-1}I$ is an isomorphism of $R$-modules.

Next, I claim that $K/S^{-1}I$ is the injective envelope of $S^{-1}R/S^{-1}I$ viewed as an $S^{-1}R$-module. This is a generalization of the fact mentioned above: The ring $S^{-1}R$ is a PID with finitely many primes, namely, $S^{-1}P_1,\dots,S^{-1}P_t$ and $S^{-1}I$ is a nonzero ideal contained in their product. Using this, it is routine to check that $S^{-1}R/S^{-1}I$ is an essential $S^{-1}R$-submodule of $K/S^{-1}I$, and the latter is injective over $S^{-1}R$ because it is divisible.

To finish the proof it is enough to show that if $M$ is an $R$-module such that $S^{-1}M$ is injective as an $S^{-1}R$-module, then $S^{-1}M$ is injective as an $R$-module. This follows by writing down the diagram definition of injectivity, noting that the diagram maps into its localization relative to $S$, and applying the injectivity of $S^{-1}M=S^{-1}(S^{-1}M)$ over $S^{-1}R$ to the latter diagram.

**Edit.** Concerning your question about the dimension of $E(R/P)$, if $R$ is a PID containing a field $k$ and $P$ is a nonzero prime ideal such that $\dim_k (R/P)=1$ (in your question $k=\mathbb{C}$), then $\dim_k E(R/P)$ is countable.

*Proof.*
Suppose $P=pR$. As explained above, $E(R/P)=K/P_P$, where $K$ is the fraction field of $R$.
I claim that the set $\{1,p^{-1},p^{-2},\dots\}$ spans $K/P_P$ as a $k$-vector space. (In fact, it is a $k$-basis.)

Given $r\in K$, there is some positive integer $n$ such that $r\in p^{-n} R_P$.
Write $r=p^{-n}a$ with $a\in R_P$.
Since the natural map $k\to R/P\to R_P/P_P$ is an isomorphism, there is $\alpha_{-n}\in k$ such that $\alpha_n$ and $a$ have the same image in $R_P/P_P$. Thus, $a-\alpha_{-n}\in P_P=pR_P$ and we can write $r=\alpha_{-n} p^{-n}+r'$ with $r'=p^n(a-\alpha_{-n})\in p^{-n+1} R_P$. Applying the same argument to $r'$, we see that there is $\alpha_{-n+1}\in k$ such that $r=\alpha_{-n} p^{-n}+\alpha_{-n+1}p^{-n+1} +r''$ with $r''\in p^{-n+2}R_P$. Proceeding by induction, we eventually find that
$$r=\alpha_{-n} p^{-n}+\alpha_{-n+1}p^{-n+1} +\dots+ \alpha_0p^0+r_1$$ with $r_1\in P_P$. Thus, $r\equiv \alpha_{-n} p^{-n}+\alpha_{-n+1}p^{-n+1} +\dots+ \alpha_0p^0\bmod P_P$, which is what we want.

*Remark.*
One can elaborate this argument further to show that $E(R/P)=K/P_P$ can be described as the set of formal power series
$$
\alpha_{-\ell} p^{-\ell}+\dots+\alpha_0 p^0
$$
with $\alpha_0,\dots,\alpha_{-\ell}\in k$ and $r\in\mathbb{N}$.
The action of $R$ is given as follows: Given $r\in R$, write it as $r=\beta_0+\beta_1p+\dots+\beta_t p^t+r_{t+1}$ with $r_{t+1}\in p^{t+1}R$ and $\beta_0,\dots,\beta_t\in k$ for $t$ sufficiently large (i.e. $t>\ell$). Compute the formal product $(\alpha_{-\ell} p^{-\ell}+\dots+\alpha_0 p^0)(\beta_0+\beta_1p+\dots+\beta_t p^t)$ and truncate the positive $p$-powers.