[proofplan] Choose a maximal subset $X \subset A$ whose translates $x+B$ are pairwise disjoint. The disjointness forces $|X||B| \le |A+B|$, and the hypothesis then gives $|X| \le K$. Maximality says that every translate $a+B$ with $a \in A$ must intersect one of the chosen translates, and this intersection rewrites $a$ as an element of $X+B-B$. [/proofplan] [step:Choose a maximal disjoint family of translates of $B$ inside $A+B$] For each $a \in A$, define the translate \begin{align*} a+B := \{a+b : b \in B\} \subset G. \end{align*} Since $A$ is finite, there exists a subset $X \subset A$ maximal with respect to the property that the family of sets \begin{align*} \{x+B : x \in X\} \end{align*} is pairwise disjoint. Here maximal means that no element $a \in A \setminus X$ can be added to $X$ while preserving pairwise disjointness of the translates. [/step] [step:Bound the size of the maximal set by comparing disjoint unions] For every $x \in X$, we have $x+B \subset A+B$ because $x \in X \subset A$. The translates $x+B$ are pairwise disjoint by the choice of $X$, and each translate has cardinality $|B|$ because the map \begin{align*} B &\to x+B \\ b &\mapsto x+b \end{align*} is a bijection with inverse $y \mapsto y-x$. Therefore \begin{align*} |X||B| = \left|\bigcup_{x \in X} (x+B)\right| \le |A+B| \le K|B|. \end{align*} Since $B$ is nonempty, $|B|>0$, so division by $|B|$ gives \begin{align*} |X| \le K. \end{align*} [/step] [step:Use maximality to cover every element of $A$ by $X+B-B$] Let $a \in A$. If $a \in X$, then since $B$ is nonempty, choose $b_0 \in B$ and write \begin{align*} a = a+b_0-b_0 \in X+B-B. \end{align*} Now suppose $a \in A \setminus X$. By maximality of $X$, adding $a$ to $X$ would destroy pairwise disjointness. Hence there exists $x \in X$ such that \begin{align*} (a+B) \cap (x+B) \ne \varnothing. \end{align*} Choose an element $y \in (a+B) \cap (x+B)$. Then there exist $b_1,b_2 \in B$ such that \begin{align*} y = a+b_1 = x+b_2. \end{align*} Rearranging in the abelian group $G$ gives \begin{align*} a = x+b_2-b_1. \end{align*} Since $x \in X$ and $b_2,b_1 \in B$, this shows $a \in X+B-B$. [/step] [step:Conclude the covering inclusion] The previous step proves that every $a \in A$ belongs to $X+B-B$. Therefore \begin{align*} A \subset X+B-B. \end{align*} Together with the bound $|X| \le K$, this is the desired conclusion. [/step]