The vectors $\mathbf{a}$ and $\mathbf{b}$ are given by
$\mathbf{a} = 4\mathbf{i} + 6p\mathbf{j} - 8\mathbf{k} \quad \text{and} \quad \mathbf{b} = 2\mathbf{i} - 3\mathbf{j} + 4p\mathbf{k} \text{, where } p > 0.$ It is given that $|\mathbf{a}| = 2|\mathbf{b}|$.

1. Express the magnitude squared for both vectors $\mathbf{a}$ and $\mathbf{b}$

$|\mathbf{a}|^2 = 4^2 + (6p)^2 + (-8)^2 = 16 + 36p^2 + 64 = 80 + 36p^2$
$|\mathbf{b}|^2 = 2^2 + (-3)^2 + (4p)^2 = 4 + 9 + 16p^2 = 13 + 16p^2$

2. Form an equation using the given relationship and solve for $p$

Given $|\mathbf{a}| = 2|\mathbf{b}|$, we can square both sides to get $|\mathbf{a}|^2 = 4|\mathbf{b}|^2$.
$80 + 36p^2 = 4(13 + 16p^2)$
$80 + 36p^2 = 52 + 64p^2$
$28p^2 = 28$
$p^2 = 1$
Since $p > 0$, we have $p = 1$.

1. State the geometrical interpretation using the properties of the scalar product

We can rewrite the expression as $\left| \mathbf{a} \cdot \frac{\mathbf{b}}{|\mathbf{b}|} \right| = |\mathbf{a} \cdot \hat{\mathbf{b}}|$.
Since $\hat{\mathbf{b}}$ is the unit vector in the direction of $\mathbf{b}$, $|\mathbf{a} \cdot \hat{\mathbf{b}}|$ represents the length of the projection of $\mathbf{a}$ onto $\mathbf{b}$.

1. Determine the value of $p$ from part (i) and substitute into the vectors.

$|\mathbf{a}|^2 = 4|\mathbf{b}|^2 \implies 16 + 36p^2 + 64 = 4(4 + 9 + 16p^2)$
$80 + 36p^2 = 52 + 64p^2 \implies 28p^2 = 28$
Since $p > 0$, $p = 1$.
Thus, $\mathbf{a} = \begin{pmatrix} 4 \\ 6 \\ -8 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} 2 \\ -3 \\ 4 \end{pmatrix}$.

2. Calculate the scalar product $\mathbf{b} \cdot \mathbf{a}$.

$\mathbf{b} \cdot \mathbf{a} = \begin{pmatrix} 2 \\ -3 \\ 4 \end{pmatrix} \cdot \begin{pmatrix} 4 \\ 6 \\ -8 \end{pmatrix} = (2)(4) + (-3)(6) + (4)(-8)$
$= 8 - 18 - 32 = -42$

3. Calculate the magnitude of $\mathbf{b}$ and evaluate the required expression.

$|\mathbf{b}| = \sqrt{2^2 + (-3)^2 + 4^2} = \sqrt{4 + 9 + 16} = \sqrt{29}$
$\frac{1}{|\mathbf{b}|} |\mathbf{b} \cdot \mathbf{a}| = \frac{|-42|}{\sqrt{29}} = \frac{42}{\sqrt{29}} = \frac{42\sqrt{29}}{29}$