A particle moving along a path at time $t$, where $0 < t < \dfrac{\pi}{3}$, is defined parametrically by $x = \cot 3t$ and $y = 2\,\cosec 3t + 1$.

1. Differentiate the parametric equations for $x$ and $y$ with respect to $t$.

Given $x = \cot 3t$ and $y = 2\cosec 3t + 1$:
$\frac{dx}{dt} = -3\cosec^2 3t$
$\frac{dy}{dt} = 2(-3\cosec 3t \cot 3t) = -6\cosec 3t \cot 3t$

2. Apply the chain rule to find the derivative $\frac{dy}{dx}$ in terms of $t$.

$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{-6\cosec 3t \cot 3t}{-3\cosec^2 3t}$
$\frac{dy}{dx} = \frac{2\cot 3t}{\cosec 3t} = \frac{2\left(\frac{\cos 3t}{\sin 3t}\right)}{\frac{1}{\sin 3t}} = 2\cos 3t$

3. Form the equation of the tangent at point $P$ and substitute $x = 0$ to find the y-intercept.

At $t = p$, the gradient of the tangent is $m = 2\cos 3p$.
The equation of the tangent at $P(\cot 3p,\; 2\cosec 3p + 1)$ is:
$y - (2\cosec 3p + 1) = 2\cos 3p (x - \cot 3p)$
To find where it meets the y-axis, set $x = 0$:
$y = 2\cosec 3p + 1 + 2\cos 3p(0 - \cot 3p)$
$y = 2\cosec 3p + 1 - 2\cos 3p \cot 3p$

4. Simplify the expression for the y-coordinate using trigonometric identities to show the required form.

Expressing in terms of sine and cosine:
$y = \frac{2}{\sin 3p} + 1 - 2\cos 3p \left(\frac{\cos 3p}{\sin 3p}\right)$
$y = \frac{2 - 2\cos^2 3p}{\sin 3p} + 1$
$y = \frac{2(1 - \cos^2 3p)}{\sin 3p} + 1$
$y = \frac{2\sin^2 3p}{\sin 3p} + 1$
$y = 2\sin 3p + 1$
Thus, the coordinates of $Q$ are $(0,\; 2\sin 3p + 1)$.

1. Express the square of the distance $s^2$ in terms of the parameter $t$ and simplify.

$s^2 = x^2 + (y - 1)^2$
$s^2 = (\cot 3t)^2 + (2\csc 3t + 1 - 1)^2$
$s^2 = \cot^2 3t + 4\csc^2 3t$
Using the identity $\cot^2 \theta = \csc^2 \theta - 1$:
$s^2 = (\csc^2 3t - 1) + 4\csc^2 3t$
$s^2 = 5\csc^2 3t - 1$

2. Differentiate the expression for $s^2$ implicitly with respect to $t$.

$\frac{\mathrm{d}}{\mathrm{d}t}(s^2) = \frac{\mathrm{d}}{\mathrm{d}t}(5\csc^2 3t - 1)$
$2s \frac{\mathrm{d}s}{\mathrm{d}t} = 5(2\csc 3t)(-3\csc 3t \cot 3t)$
$2s \frac{\mathrm{d}s}{\mathrm{d}t} = -30\csc^2 3t \cot 3t$

3. Evaluate $s$ and $\frac{\mathrm{d}s}{\mathrm{d}t}$ at $t = \frac{\pi}{4}$.

When $t = \frac{\pi}{4}$, $3t = \frac{3\pi}{4}$.
$\csc\left(\frac{3\pi}{4}\right) = \sqrt{2} \implies \csc^2\left(\frac{3\pi}{4}\right) = 2$
$\cot\left(\frac{3\pi}{4}\right) = -1$
Substitute these into the equation for $s^2$:
$s^2 = 5(2) - 1 = 9 \implies s = 3 \quad (\text{since } s > 0)$
Substitute the values into the derivative equation to find $\frac{\mathrm{d}s}{\mathrm{d}t}$:
$2(3) \frac{\mathrm{d}s}{\mathrm{d}t} = -30(2)(-1)$
$6 \frac{\mathrm{d}s}{\mathrm{d}t} = 60$
$\frac{\mathrm{d}s}{\mathrm{d}t} = 10$