(i) To find the equation of the normal, first find the derivative of \(y = \sqrt{5x - 1}\). The derivative is \(\frac{dy}{dx} = \frac{1}{2}(5x - 1)^{-\frac{1}{2}} \times 5 = \frac{5}{2\sqrt{5x - 1}}\).

At \(P(2, 3)\), \(\frac{dy}{dx} = \frac{5}{2 \times 3} = \frac{5}{6}\).

The slope of the normal is the negative reciprocal: \(m = -\frac{6}{5}\).

The equation of the normal is \(y - 3 = -\frac{6}{5}(x - 2)\).

(ii) To find the area of the shaded region, integrate the curve from \(x = \frac{1}{5}\) to \(x = 2\):

\(\int_{\frac{1}{5}}^{2} \sqrt{5x - 1} \, dx = \left[ \frac{2}{15}(5x - 1)^{\frac{3}{2}} \right]_{\frac{1}{5}}^{2} = 3.6\).

The normal crosses the x-axis at \(x = \frac{27}{5}\).

The area of the triangle formed by the normal and the x-axis is \(\frac{1}{2} \times \frac{15}{4} \times 3 = 3.75\).

Total area of the shaded region is \(3.6 + 3.75 = 7.35\) or \(\frac{147}{20}\).