$$\boxed{\sqrt{25+12\sqrt{3}}}$$ Because $3, 4$, and $5$ form a Pythagorean triplet. Let's try to construct a right triangle by the rotation method\footnote{The rotation method is discussed in the book \textit{Geometry Techniques} by the same author}. In order to achieve this, let's rotate $\triangle{ABP}$ around point $A$ for $60^\circ$ counterclockwise. Because $\triangle{ABC}$ is equilateral, point $B$ will move to point $C$. Let point $P$'s new location be $P'$. See the diagram $(i)$ below.

Now we find $AP'=AP=3$ and $\angle{PAP'}=60^\circ$, therefore $\triangle{APP'}$ is equilateral. This implies $PP'=3$ too. Obviously, $CP'=BP=4$ and $CP=5$ Therefore $CPP'$ is a right triangle by the converse of Pythagorean theorem where $\angle{CP'P} = 90^\circ$. Next, draw $AE\perp PP'$ and let the foot be $E$. See the diagram $(ii)$ above. $$\angle{AP'E} = 180^\circ - \angle{PP'C} - \angle{AP'P} = 180^\circ - 90^\circ - 60^\circ = 30^\circ$$ Therefore $$AE=\frac{1}{2}AP' = \frac{3}{2}\quad\text{and}\quad CE = \frac{\sqrt{3}}{2}AE+CP'=\frac{3\sqrt{3}}{2}+4$$ Applying Pythagorean theorem on $\triangle{AEC}$ yields: $$AC=\sqrt{\Big(\frac{3}{2}\Big)^2 + \Big(\frac{3\sqrt{3}}{2}+4\Big)^2}=\sqrt{25+12\sqrt{3}}$$