Since \(P\) lies in the third quadrant, \(y\) must be negative. Also, because \(P\) lies on the unit circle, its coordinates must satsify
\(\left(-\dfrac35\right)^2+y^2=1\implies y^2=\dfrac{16}{25}\implies y=\pm\dfrac45\)
Take the negative root, since \(y<0\), so that \(y=-\dfrac45\).
