Why is log(n/p) asymptotically less than log(n)/log(p)
I assume that $p &gt; 1$ (which is equivalent to $\log p &gt; 0$, otherwise we have negative-valued functions which are meaningless as complexity measures.

For a given $p$,  $O\left(\frac{\log n}{\log p}\right) = O(\log n)$ since big oh isn't affected by multiplication by a positive constant.
$\log \frac{n}{p} = \log n - \log p$. Since $\log p = o(\log n)$, $O\left(\log \frac{n}{p}\right) = O(\log n)$.

So up to big-oh complexity (or big-theta, for the same reasons), these two classes are the same.

Your diagram doesn't show a visual difference: you need to bring the two curves to the same scale. One is $a \log n$ for some constant $a &gt; 0$, the other is $\log n + b$ for some constant $b$.

If you want to study the variation in $p$, then $O$ or $\Theta$ with respect to the variable $n$ are not good ways of modeling the complexity of your problem. You need more precise approximations that treat multiplicative constants as relevant. The complexity in $p$ is probably relevant to your problem, and is very different. I'm not familiar enouhg with parallelism to know whether it's the right measure to use here, this could warrant a separate question (with more information about what you're modeling). You're going to need express the complexity in terms of $n$ and $p$, and look at how the variation in $p$ (not the complexity itself) behaves for large $n$.