Is the following map $i$ a complete embedding or a dense embedding from the Prikry-Silver forcing $\mathbb P\mathbb S$ to the Sacks forcing $\mathbb S$? \begin{align} i \colon \mathbb P\mathbb S \to \mathbb S: p \mapsto \{s : s \in 2^{<\omega}, \text{$p \cup s$ is a function}\} \end{align} where $\mathbb S := \{T : \text{$T$ is a subtree of $2^{<\omega}$}, \text{$T$ is perfect}\}$, $\le_{\mathbb S} := \subseteq$, $\mathbb P\mathbb S := \{p : \text{$p$ is a function}, \mathrm{dom}(p) \subseteq \omega, \mathrm{ran}(p) \subseteq 2, \lvert \omega \setminus \mathrm{dom}(p)\rvert = \aleph_0 \}$, $\le_{\mathbb P\mathbb S} := \supseteq$
Any contribution, thank you very much.
No. Here's an "abstract nonsense" style argument, at least assuming CH. Both Sacks and Prikry–Silver reals are minimal, but Sacks reals also have the property that for any new real is actually generic for the Sacks forcing (namely, there's an automorphism which maps the canonical Sacks real to a name for that fixed new real).
So, if the map was a complete embedding, it would mean that any Sacks real adds a Prikry–Silver real is actually a Sacks real in disguise. Which is something we know isn't true.
More concretely, though, note that a condition in the Prikry–Silver forcing translates to a "uniformly perfect" tree, where for any $|s|=|t|$, both in $T$, the extension $s^\frown i$ is in $T$ if and only if $t^\frown i\in T$, for $i\in\{0,1\}$.
Now, start with a perfect tree and recursively prune it in a way that is non-uniformly perfect. That is not very hard, and it shows more directly that this embedding is not dense and not complete.