(i) The gradient of line \(AP\) is given by:

\(\frac{b - 1}{a + 1} = 2\)

Solving for \(b\):

\(b - 1 = 2(a + 1)\) \(b - 1 = 2a + 2\) \(b = 2a + 3\)

Thus, \(b = 2a + 3\).

(ii) The distance \(AB\) is:

\(AB = \sqrt{(10 - (-1))^2 + (-1 - 1)^2} = \sqrt{11^2 + 2^2} = \sqrt{125}\)

Since \(AP = AB\), we have:

\(\sqrt{(a + 1)^2 + (b - 1)^2} = \sqrt{125}\)

Substitute \(b = 2a + 3\):

\((a + 1)^2 + (2a + 3 - 1)^2 = 125\) \((a + 1)^2 + (2a + 2)^2 = 125\) \((a + 1)^2 + 4(a + 1)^2 = 125\) \(5(a + 1)^2 = 125\) \((a + 1)^2 = 25\) \(a + 1 = 5 \quad \text{or} \quad a + 1 = -5\) \(a = 4 \quad \text{or} \quad a = -6\)

For \(a = 4\):

\(b = 2(4) + 3 = 11\)

For \(a = -6\):

\(b = 2(-6) + 3 = -9\)

Thus, the possible coordinates of \(P\) are \((4, 11)\) and \((-6, -9)\).