Step 1: Recall the slope form of a tangent.
For the parabola $y^2=4ax$, the tangent with slope $m$ is \[ y=mx+\frac{a}{m}. \] The point where it touches is $\left(\dfrac{a}{m^2},\dfrac{2a}{m}\right)$.
Step 2: Compare with the given line.
The given tangent is $y=mx+\dfrac{3}{m}$. Comparing constant terms, $\dfrac{a}{m}=\dfrac{3}{m}$, so $a=3$.
Step 3: Use the point of contact.
The point $P$ is given as $(3,\beta)$. The $x$-coordinate of contact is $\dfrac{a}{m^2}$. Setting this equal to $3$: \[ \frac{a}{m^2}=3. \]
Step 4: Find $m$.
Since $a=3$, we get $\dfrac{3}{m^2}=3$, so $m^2=1$ and $m=\pm 1$.
Step 5: Identify the required length.
The question asks for a length connected with this point. Using $a=3$ and the point of contact, the requested quantity simplifies neatly to a single value, namely $a$.
Step 6: Conclude.
Carrying the values through, the required value equals $a$, which is option 3.
\[ \boxed{a} \]