Is finding a minimal set of seed variables for a complete deduction of a system of equations NP-complete?
Yes, it is $NP$-complete even if $d=4$.

Note that unless $P=NP$, we cannot prove $NP$-completeness for $d=2$ because it is merely  the problem of counting the number of connected components of a given graph.

Reduce from Exact-Cover-by-$3$-sets (X3C).

Given an instance of X3C, we construct an instance of our problem as follows.

The set of elements is $E=\{e_1,e_2,\dots,e_n\}$.
The set of $3$-sets is $C=\{s_1,s_2,\dots,s_m\}$.

This will be a YES instance (of X3C) iff there exists $C'\subseteq C$ with $|C'|=\frac n3$ that cover each element exactly once.

Now, set the set of variables $V=C\cup E\cup C^{up}$ where $C^{up}=\{s_j^{up}|s_j\in C\}$, that means each element is a varialbe and each $3$-set is also a variable. And for each set $s_j$, we add a new variable $s_j^{up}$.

Note that, X3C remains $NP$-complete even when each element belongs to at most $2$ sets and each pair of $3$-sets shares at most one element.

So, here come the equations.

For each $3$-set $s_j\in C$ where $s_j=\{e_a,e_b,e_c\}$, add the following equations:


$\{s_j,e_a,e_b,e_c\}$
$\{s_j,s_j^{up}\}$
$\{s_j,s_j^{up},e_a\}, \{s_j,s_j^{up},e_b\}, \{s_j,s_j^{up},e_c\}$


Finally, set $k=\frac n3$.

We have done with our reduction. Now we have to prove its correctness.

Proof of correctness: (sketchy)

For one direction, if we have a solution $C'$ to our given X3C, it is obvious that $C'$ is also a solution to our instance.

For the other direction, note that it is w.l.o.g to assume that a (minimal) solution contains only variable in $C$. Because if it contains an element (in $E$), we can safely remove it from the solution and add one of the containing $3$-sets (that the element belongs to) to the current solution (if that set has not already been present in the current solution).

So, now a subcollection $C'\subseteq C$ is a solution to our problem. We show that it is also a solution to the original X3C instance.

For each $s_j\notin C'$, it must be the case that all $3$ contained elements of $s_j$ must be deducible. For otherwise, $s_k$ becomes unreachable.

For each element $e_i\in E$, it must be the case that one of the $2$ sets containing it must be included in $C'$.$\blacksquare$



Take @xskxzr's example: The given X3C instance is $E=\{e_1,\dots,e_9\}$, $C=\{s_1,\dots,s_5\}$ where $s_1=\{e_1,e_2,e_3\}$, $s_2=\{e_4,e_5,e_6\}$, $s_3=\{e_2,e_5,e_8\}$, $s_4=\{e_6,e_7,e_9\}$, $s_5=\{e_1,e_4,e_7\}$.

Our set of variables will be $V=C\cup C^{up}\cup E=\{s_1,\dots,s_5,s_1^{up},\cdots,s_5^{up},e_1,\dots,e_9\}$.

Our set of equations will be:


$\{s_1,s_1^{up}\},\dots,\{s_5,s_5^{up}\}$
$\{s_1,e_1,e_2,e_3\}$, $\{s_2,e_4,e_5,e_6\}$, $\{s_3,e_2,e_5,e_8\}$, $\{s_4,e_6,e_7,e_9\}$, $\{s_5,e_1,e_4,e_7\}$
$\{s_1,s_1^{up},e_1\},\{s_1,s_1^{up},e_2\},\{s_1,s_1^{up},e_3\}$
$\{s_2,s_2^{up},e_4\},\{s_2,s_2^{up},e_5\},\{s_2,s_2^{up},e_6\}$
$\{s_3,s_3^{up},e_2\},\{s_3,s_3^{up},e_5\},\{s_3,s_3^{up},e_8\}$
$\{s_4,s_4^{up},e_6\},\{s_4,s_4^{up},e_7\},\{s_4,s_4^{up},e_9\}$
$\{s_5,s_5^{up},e_1\},\{s_5,s_5^{up},e_4\},\{s_5,s_5^{up},e_7\}$


And $k=\frac n3=3$.