Let $P$ be a prime ideal of $\mathbb C[x_1,...,x_d]$ such that ht$(P)=d-1$ i.e. $\dim (\mathbb C[x_1,...,x_d]/P)=1$.
Then is it necessarily true that there exists an injective $\mathbb C$-algebra homomorphism $\mathbb C[x_1,...,x_d]/P\to \mathbb C[[T]]$ ? If this is not true in general, would it be true if I moreover assumed $P$ is a homogeneous prime ideal ?
The prime ideal $P$ corresponds to a curve $X \subseteq \Bbb P^d_{\Bbb C}$, so that the ring $\Bbb C[x_1, \dotsc, x_d]/P$ is just the coordinate ring $\Bbb C[X]$.
If you choose any smooth $\Bbb C$-point $x$ on $X$ (which exists, because the set of smooth points are Zariski dense), then the local ring $\Bbb C[X]_x$ has formal completion isomorphic to $\Bbb C[[T]]$. This gives an embedding $\Bbb C[X] \rightarrow \Bbb C[[T]]$.
One intuitive way of seeing this is to choose a local coordinate $T$ at the smooth point $x$, so that any rational function $f \in \Bbb C[X]$ can be viewed as an analytic function defined on an open neighborhood (in the usual analytic topology) of $T = 0$ in $\Bbb C$, and hence has a Taylor expansion (i.e. Laurent expansion without negative terms). Two analytic functions having the same Taylor expansion must be identical on the open neighborhood, and hence identical on the whole $X$, by irreducibility.