If $z=\sqrt{2}\sqrt{1+\sqrt{3}i}$  represents a point ‘P’ in the argand plane and ‘P’ lies in the third quadrant then the polar form of z is 

$\begin{array}{rl}z& =\sqrt{2}\sqrt{1+\sqrt{3}i}\\ z& =\sqrt{2+2\sqrt{3}i}\\ & =\sqrt{(\sqrt{3}+i{)}^2}\\ & =\pm (\sqrt{3}+i)\end{array}$
Since point P lies in third quadrant then $z=-\sqrt{3}-i$
Now  polar form of z is $2[-\frac{\sqrt{3}}{2}+i\left(\frac{-1}{2}\right)]$
$=2[\cos \left(\frac{-5\pi }{6}\right)+i\sin \left(\frac{-5\pi }{6}\right)]$