 Solution

Part c gives $x+B(\Gamma ,\alpha )\subseteq A+A+A$ with $d=|\Gamma |\le 2{\alpha }^{-3}$. By the arithmetic progression in a cyclic Bohr set, $B(\Gamma ,\alpha )$ contains a centered progression of length at leastIts translate by $x$ is the required arithmetic progression in $A+A+A$.