To find the stationary point, we first differentiate the equation \(y = ax^{\frac{1}{2}} - 2x\) with respect to \(x\):

\(\frac{dy}{dx} = \frac{1}{2}ax^{-\frac{1}{2}} - 2\).

At a stationary point, \(\frac{dy}{dx} = 0\). Substituting \(x = 9\) into the derivative:

\(0 = \frac{1}{2}a(9)^{-\frac{1}{2}} - 2\).

This simplifies to:

\(0 = \frac{a}{6} - 2\).

Solving for \(a\):

\(\frac{a}{6} = 2\)

\(a = 12\).

Now, substitute \(a = 12\) back into the original equation to find the \(y\)-coordinate of \(P\):

\(y = 12 \times 9^{\frac{1}{2}} - 2 \times 9\).

\(y = 12 \times 3 - 18\).

\(y = 36 - 18\).

\(y = 18\).