(a) To find the equation of line \(BC\), we use the point-slope form. The slope \(m\) of \(BC\) is calculated using the coordinates of \(B(0, 2)\) and the center \(C(2, -4)\):

\(m = \frac{-4 - 2}{2 - 0} = -3\)

Using the point \(B(0, 2)\), the equation of the line is:

\(y - 2 = -3(x - 0)\)

Simplifying gives:

\(y = 2 - 3x\)

(b) To find the coordinates of \(P\), substitute the equation of \(BC\) into the circle's equation:

\((x-2)^2 + (2 - 3x + 4)^2 = 20\)

\((x-2)^2 + (6 - 3x)^2 = 20\)

Expanding and simplifying:

\((x-2)^2 + (6 - 3x)^2 = 20\)

\((x-2)^2 + (36 - 36x + 9x^2) = 20\)

\(10x^2 - 40x + 20 = 0\)

Solving the quadratic equation:

\(x = 2 \pm \sqrt{2}\)

For \(x = 2 - \sqrt{2}\), substitute back into \(y = 2 - 3x\):

\(y = 2 - 3(2 - \sqrt{2}) = 3\sqrt{2} - 4\)

Thus, the coordinates of \(P\) are \((2 - \sqrt{2}, 3\sqrt{2} - 4)\).