(i) To find the value of \(c\), we need the line \(4y = x + c\) to be tangent to the curve \(y^2 = x + 3\). Substitute \(x = 4y - c\) into the curve equation:

\(y^2 = 4y - c + 3\)

\(y^2 - 4y + c - 3 = 0\)

This is a quadratic in \(y\). For the line to be tangent, the discriminant must be zero:

\(b^2 - 4ac = 0\)

Here, \(a = 1\), \(b = -4\), \(c = c - 3\). So:

\((-4)^2 - 4(1)(c - 3) = 0\)

\(16 - 4c + 12 = 0\)

\(28 - 4c = 0\)

\(4c = 28\)

\(c = 7\)

(ii) With \(c = 7\), substitute back to find \(P\):

\(y^2 = 4y - 7 + 3\)

\(y^2 - 4y + 4 = 0\)

\((y - 2)^2 = 0\)

\(y = 2\)

Substitute \(y = 2\) into \(x = 4y - c\):

\(x = 4(2) - 7 = 1\)

Coordinates of \(P\) are \((1, 2)\).