1. Use the fact that $P(6,3)$ lies on the hyperbola

   The hyperbola is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1.$

   Since $P(6,3)$ lies on it, $\frac{36}{a^2}-\frac{9}{b^2}=1. \qquad (1)$

2. Find the slope of the normal at $P$

   The normal at $P$ passes through $(6,3)$ and intersects the $x$-axis at $(9,0)$.

   So its slope is $m_n=\frac{0-3}{9-6}=-1.$

   Therefore the slope of the tangent is the negative reciprocal: $m_t=1.$

3. Differentiate the hyperbola to get tangent slope

   From $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,$ differentiating implicitly: $\frac{2x}{a^2}-\frac{2y}{b^2}\frac{dy}{dx}=0$ $\Rightarrow \frac{dy}{dx}=\frac{b^2x}{a^2y}.$

   At $P(6,3)$, $m_t=\frac{b^2\cdot 6}{a^2\cdot 3}=\frac{2b^2}{a^2}.$

   But from Step 2, $m_t=1$. Hence, $\frac{2b^2}{a^2}=1$ $\Rightarrow a^2=2b^2. \qquad (2)$

4. Use point condition to find relation

   Substitute $a^2=2b^2$ into (1): $\frac{36}{2b^2}-\frac{9}{b^2}=1$ $\frac{18-9}{b^2}=1$ $\frac{9}{b^2}=1$ $b^2=9.$

   Then $a^2=2b^2=18.$

5. Find eccentricity

   For the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,$ eccentricity is $e=\sqrt{1+\frac{b^2}{a^2}}.$

   So $e=\sqrt{1+\frac{9}{18}}=\sqrt{1+\frac12}=\sqrt{\frac32}.$

6. Check options

   $\sqrt{\frac32}$ matches Option B.

Final Answer: $\boxed{\sqrt{\frac{3}{2}}}$