(i) To express \(f(x) = x^2 - 4x + k\) in the form \((x + a)^2 + b + k\), complete the square:

\(x^2 - 4x = (x-2)^2 - 4\)

Thus, \(f(x) = (x-2)^2 - 4 + k\), where \(a = -2\) and \(b = -4\).

(ii) The range of \(f\) is determined by the vertex of the parabola, which is at \(x = 2\). The minimum value of \(f(x)\) is \(-4 + k\), so \(f(x) > k - 4\) or \([k-4, \infty)\).

(iii) For \(f\) to be one-one, it must be strictly increasing or decreasing. The vertex is at \(x = 2\), so the smallest value of \(p\) is \(2\).

(iv) To find \(f^{-1}(x)\), start with \(y = (x-2)^2 - 4 + k\).

\(y - k + 4 = (x-2)^2\)

\(x - 2 = \pm \sqrt{y + 4 - k}\)

Since \(x \geq 2\), take the positive root: \(x = 2 + \sqrt{y + 4 - k}\)

Thus, \(f^{-1}(x) = 2 + \sqrt{x + 4 - k}\).

The domain of \(f^{-1}\) is \(x > k - 4\) or \([k-4, \infty)\).