Answer
$P(\frac{-\sqrt{2}}{3},\frac{-\sqrt{7}}{3})$
Work Step by Step
We have the following equation for the unit circle:
$x^{2}+y^{2}=1$, so we plug in $P(\frac{-\sqrt{2}}{3},y)$,
$(\frac{-\sqrt{2}}{3})^{2}+y^{2}=1$
$ y^{2}=1 - \frac{2}{9}= \frac{7}{9}$
$y = \frac{\sqrt{7}}{3}$or $y = - \frac{\sqrt{7}}{3}$ but P lies below the x-axis so the y-coordinate is negative so $y = - \frac{\sqrt{7}}{3}$