Step 1: Spot the derivative pattern.
In $\displaystyle\int \dfrac{\cos x - \sin x}{8 - \sin 2x}dx$, the numerator $\cos x - \sin x$ is the derivative of $\sin x + \cos x$.
Step 2: Substitute.
Let $t = \sin x + \cos x$, so $dt = (\cos x - \sin x)dx$.
Step 3: Rewrite the denominator.
$t^2 = 1 + \sin 2x$, hence $\sin 2x = t^2 - 1$, and $8 - \sin 2x = 8 - (t^2 - 1) = 9 - t^2$.
Step 4: Transform the integral.
It becomes $\displaystyle\int \dfrac{dt}{9 - t^2} = \int \dfrac{dt}{3^2 - t^2}$.
Step 5: Apply the standard formula.
$\displaystyle\int \dfrac{dt}{a^2 - t^2} = \dfrac{1}{2a}\log\left|\dfrac{a+t}{a-t}\right| + c$ with $a = 3$ gives $\dfrac{1}{6}\log\left|\dfrac{3 + \sin x + \cos x}{3 - \sin x - \cos x}\right| + c$.
Step 6: Match the parameter.
Comparing with $\dfrac{1}{p}\log[\dots]$ gives $\dfrac{1}{p} = \dfrac{1}{6}$, so $p = 6$.
\[ \boxed{p = 6} \]