= Solution
Choose one representative from $A$ above each point of $\pi(A)$ and collect them in $X$. Part (c) gives $|X|\leq K^{O(1)}$. If $a\in A$ has the same image as $x\in X$, then $x^{-1}a\in A^2\cap N$, so
$$
A\subseteq X(A^2\cap N).
$$
This is the required covering of $A$ by at most $K^{O(1)}$ left cosets of the abelian translation subgroup $N$.
The set $B=A^2\cap N$ is a $K^{O(1)}$-approximate group by the <intersection of an approximate group power with a subgroup>. Apply the <Freiman-Green-Ruzsa theorem> inside $N\cong(\mathbb C,+)$. Because the additive group of the <complex numbers> is a <torsion-free group>, the finite subgroup part is trivial, so there is an <abelian progression> $P$ with
$$
B\subseteq P,\qquad
\operatorname{rank}P\leq K^{O(1)},\qquad
|P|\leq\exp(K^{O(1)})|B|.
$$
Since $|B|\leq|A^2|\leq K|A|$, enlarging the implicit constant gives
$$
\boxed{A\subseteq XP,\qquad |X|\leq K^{O(1)},\qquad
\operatorname{rank}P\leq K^{O(1)},\qquad
|P|\leq\exp(K^{O(1)})|A|.}
$$
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