Step 1: Set up the curve.
The curve is $\dfrac{x^2}{3}+\dfrac{y^2}{27}=4$ and the point is $P(1,3\sqrt3)$. We want the length of the tangent from $P$ to where it meets the $x$-axis.
Step 2: Find the slope by implicit differentiation.
Differentiating, $\dfrac{2x}{3}+\dfrac{2y}{27}\dfrac{dy}{dx}=0$, so $\dfrac{dy}{dx}=-\dfrac{9x}{y}$.
Step 3: Evaluate at $P$.
\[ \left.\frac{dy}{dx}\right|_{(1,3\sqrt3)}=-\frac{9(1)}{3\sqrt3}=-\frac{3}{\sqrt3}=-\sqrt3. \]
Step 4: Write the tangent line.
$y-3\sqrt3=-\sqrt3(x-1)$ simplifies to $\sqrt3\,x+y=4\sqrt3$.
Step 5: Find the $x$-intercept.
Set $y=0$: $\sqrt3\,x=4\sqrt3$, so $x=4$. The tangent meets the $x$-axis at $A(4,0)$.
Step 6: Measure $PA$.
\[ PA=\sqrt{(4-1)^2+(0-3\sqrt3)^2}=\sqrt{9+27}=\sqrt{36}=6. \]
\[ \boxed{6} \]