The trajectory equation of a projectile is given by $$y = x \tan\theta - \frac{gx^2}{2v^2\cos^2\theta}$$, and the position of maximum height occurs at $$x_H = \frac{R}{2} = \frac{v^2 \sin 2\theta}{2g}$$.

Given coordinates of point $$P$$: $$(x, y) = (5, 1)$$

Evaluating Option (A) and (B) with $$\theta = 45^\circ$$:

$$1 = 5 \tan 45^\circ - \frac{g(5)^2}{2v^2 \cos^2 45^\circ}$$

$$1 = 5 - \frac{25g}{2v^2 \left(\frac{1}{2}\right)} = 5 - \frac{25g}{v^2}$$

$$\frac{25g}{v^2} = 4 \implies v^2 = \frac{25g}{4} \implies v = \frac{5\sqrt{g}}{2}\text{ ms}^{-1}$$

$$x_H = \frac{\left(\frac{25g}{4}\right) \sin 90^\circ}{2g} = \frac{25}{8} = 3.125\text{ m}$$

$$x_H = 3.125\text{ m} < 5\text{ m} \implies \text{maximum height is reached before } P$$

Evaluating Option (C) with $$\theta = 30^\circ$$:

$$1 = 5 \tan 30^\circ - \frac{g(5)^2}{2v^2 \cos^2 30^\circ} = \frac{5}{\sqrt{3}} - \frac{25g}{2v^2 \left(\frac{3}{4}\right)}$$

$$\frac{50g}{3v^2} = \frac{5}{\sqrt{3}} - 1 \implies v^2 = \frac{50g}{3\left(\frac{5}{\sqrt{3}} - 1\right)} = \frac{50g}{5\sqrt{3} - 3}$$

$$x_H = \frac{v^2 \sin 60^\circ}{2g} = \frac{50g}{2g(5\sqrt{3} - 3)} \cdot \frac{\sqrt{3}}{2} = \frac{25\sqrt{3}}{2(5\sqrt{3} - 3)} = \frac{25}{10 - 2\sqrt{3}} \approx \frac{25}{6.536} \approx 3.82\text{ m}$$

$$x_H = 3.82\text{ m} < 5\text{ m} \implies \text{maximum height is reached before } P$$

Evaluating Option (D) with $$\tan\theta = \frac{1}{5}$$:

$$1 = 5\left(\frac{1}{5}\right) - \frac{g(5)^2}{2v^2\cos^2\theta} \implies 1 = 1 - \frac{25g}{2v^2\cos^2\theta} \implies \frac{25g}{2v^2\cos^2\theta} = 0 \implies v \to \infty$$