\[\begin{align}  & p=\frac{1}{{{\log }_{a}}(2-\sqrt{3})}+\frac{1}{{{\log }_{b}}\left( \frac{\sqrt{3}-1}{\sqrt{3}+1} \right)} \\ & p=\frac{1}{{{\log }_{a}}\left( 2-\sqrt{3}\times \frac{2+\sqrt{3}}{2+\sqrt{3}} \right)}+\frac{1}{{{\log }_{b}}\left( \frac{\sqrt{3}-1}{\sqrt{3}+1}\times \frac{\sqrt{3}+1}{\sqrt{3+1}} \right)} \\ & p=\frac{1}{{{\log }_{a}}\left( \frac{1}{2+\sqrt{3}} \right)}+\frac{1}{{{\log }_{b}}\left( \frac{2}{4+2\sqrt{3}} \right)} \\ & p=\frac{1}{{{\log }_{a}}\left( \frac{1}{2+\sqrt{3}} \right)}+\frac{1}{{{\log }_{b}}\left( \frac{1}{2+\sqrt{3}} \right)} \\ & p=\frac{1}{{{\log }_{a}}{{(2+\sqrt{3})}^{-1}}}+\frac{1}{{{\log }_{b}}{{(2+\sqrt{3})}^{-1}}} \\ & p=-\frac{1}{{{\log }_{a}}(2+\sqrt{3})}-\frac{1}{{{\log }_{b}}(2+\sqrt{3})} \\ & \text{Change the base to (2+}\sqrt{3}) \\ & p=-{{\log }_{(2+\sqrt{3})}}a-{{\log }_{(2+\sqrt{3})}}b \\ & p=-({{\log }_{(2+\sqrt{3})}}a+{{\log }_{(2+\sqrt{3})}}b) \\ & p=-lo{{g}_{(2+\sqrt{3})}}ab \\ & -p={{\log }_{(2+\sqrt{3})}}ab \\ & {{(2+\sqrt{3})}^{-p}}=ab \\ & \frac{1}{{{(2+\sqrt{3})}^{p}}}=ab \\ & \text{Since we are given that (2+}\sqrt{3}{{)}^{p}}=\frac{1}{12} \\ & ab=\frac{1}{\tfrac{1}{12}}=12 \\\end{align}\]