Solution:

Given, focus of the parabola is (0, 1) and its directrix is x = 2.

Let P (x, y) be any point on the parabola.

Let perpendicular distance of point P from directrix be d and distance of point P from focus be PF.

Then, by definition of parabola, we know that d = PF.

Let's find the equation of the parabola using the above information.

Let the equation of the parabola be y2 = 4ax.

Since the focus is (0, 1), we have a = 1/4.

Also, since the directrix is x = 2, we have the equation of the directrix as x = -a.

Substituting the value of a, we get x = -1/4.

So, the equation of the parabola is y2 = x + 1/4.

Let P (x, y) be any point on the parabola.

Then, the distance of point P from focus (0, 1) is PF = √(x2 + (y-1)2).

Also, since the directrix is x = 2, the perpendicular distance of point P from directrix is d = |x - 2|.

Since d = PF, we have √(x2 + (y-1)2) = |x - 2|.

Squaring both sides, we get x2 + (y-1)2 = (x-2)2.

Expanding, we get y2 - 2y + 1 = x2 - 4x + 4.

Substituting the value of a, we get y2 = x + 1/4.

Substituting this value in the above equation, we get x + 1/4 - 2y + 1 = x2 - 4x + 4.

Simplifying, we get x2 - 3x - 4y + 15/4 = 0.

This is the required equation of the parabola.

Now, we need to find the point P (x, y) on the parabola that satisfies the given conditions.

Substituting x = t2 - 1 and y = 2t - 1 in the above equation, we get:

(t2 - 1)2 - 3(t2 - 1) - 4(2t - 1) + 15/4 = 0.

Simplifying, we get t2 - 2t + 1 = 0.

Solving for t, we get t = 1.

Substituting t = 1 in x = t2 - 1 and y = 2t - 1, we get P (0, 1).

Therefore, the point on the parabola whose focus is (0, 1) and the directrix is x = 2 is (t2 - 1, 2t - 1) = (1-1, 2-1) = (0, 1).

Hence, the correct answer is option D.