Since \(P\) lies in the fourth quadrant, you know the sine is negative.
Also since the point lies on the unit circle, the coordinates satisfy
\(x^2+y^2=1\implies \left(\dfrac{33}{64}\right)^2+y^2=1\)
When you solve for \(y\), you have two possible solutions, which are
\(y=\pm\sqrt{1-\left(\dfrac{33}{64}\right)^2}=\pm\sqrt{\dfrac{3007}{4096}}=\pm\dfrac{\sqrt{3007}}{64}\)
but since \(y\) is negative, you take the negative root.
So,
\(\sin\theta=\dfrac yx=\dfrac{-\frac{\sqrt{3007}}{64}}{\frac{33}{64}}=-\dfrac{\sqrt{3007}}{33}\)
