You know that point

For $PA + PB + PC$ to be minimized, $P$ is a Fermat point . Since $\angle A = 60°$ and the angle sum of a triangle is $180°$ , $\angle B$ and $\angle C$ are not $\geq 120°$ , so by the properties of a Fermat point $\angle APB = \angle BPC = \angle APC = 120°$ .

Let $x = \angle PCA$ . Then by the angle sum of $\triangle ACP$ , $\angle PAC + x + 120° = 180°$ , so $\angle PAC = 60° - x$ .

Since $\angle PAC = 60° - x$ and $\angle BAC = 60°$ , $\angle BAP = x$ .

Therefore, $\triangle APC \sim \triangle BPA$ by AA similarity, so $\frac{PC}{PA} = \frac{PA}{PB}$ , or $\frac{16}{PA} = \frac{PA}{9}$ , which solves to $PA = \boxed{12}$ .

The point you are looking for is the Fermat's Point. Hence using it's properties along with the law of cosines in the four triangles $PAB, PBC, PAC$ and $ABC$ we get the answer as $PA=12$