Eccentricity of Hyperbola Given Normal Intersection Point

The equation of the given hyperbola is:
$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$

The equation of the normal to the hyperbola at any point $P(x_1, y_1)$ is given by:
$\frac{a^2 x}{x_1} + \frac{b^2 y}{y_1} = a^2 + b^2$

For the point $P(6,3)$, substituting $x_1 = 6$ and $y_1 = 3$ into the normal equation gives:
$\frac{a^2 x}{6} + \frac{b^2 y}{3} = a^2 + b^2$

It is given that this normal line intersects the $x$-axis at the point $(9,0)$. Substituting $x = 9$ and $y = 0$ into the normal equation:
$\frac{a^2 (9)}{6} + \frac{b^2 (0)}{3} = a^2 + b^2$

$\frac{3}{2}a^2 = a^2 + b^2$

$\frac{3}{2}a^2 - a^2 = b^2 \implies b^2 = \frac{a^2}{2}$

The formula for the eccentricity $e$ of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is:
$e = \sqrt{1 + \frac{b^2}{a^2}}$

Substituting $b^2 = \frac{a^2}{2}$ into the formula for $e$:
$e = \sqrt{1 + \frac{\frac{a^2}{2}}{a^2}} = \sqrt{1 + \frac{1}{2}} = \sqrt{\frac{3}{2}}$

(Note: Since $P(6,3)$ lies on the hyperbola, $\frac{36}{a^2} - \frac{9}{b^2} = 1 \implies \frac{36}{a^2} - \frac{18}{a^2} = 1 \implies a^2 = 18$ and $b^2 = 9$, which confirms that a valid hyperbola exists).

Thus, the eccentricity of the hyperbola is $\sqrt{\frac{3}{2}}$.

Hence, the correct option is B.