30-60-90

BP = CP = 4; AB = BP*cos30

$\angle CBA = 60 \\ \tan(\angle CBA) = \tan(60) = \sqrt[]{3} \\ \dfrac{AC}{AB} = \sqrt[]{3} \\ \dfrac{4+AP}{AB} = \sqrt[]{3} \\ \dfrac{4+AP}{\sqrt[]{3}} = AB ...(i)\\ \angle PBA = 30 \\ \tan(30) = \dfrac{1}{\sqrt[]{3}} \\ \dfrac{AP}{AB} = \sqrt[]{3} \\ AB = \sqrt[]{3}AP ... (ii) \\ \text{(i) and (ii)} \\ \implies \dfrac{4+AP}{\sqrt[]{3}} = \sqrt[]{3}AP\\ 4 + AP = 3AP \\ AP = 2\\ \boxed{\therefore AB = \sqrt[]{3}AP = \sqrt[]{3}2 = 2\sqrt[]{3}}$

The image shows half of an equilateral triangle with point $P$ its incenter. That makes $AP=4/2=2$ , $AB=2/\tan(30)=2\sqrt{3}$ .