Consider the following code
f[n_, p_] := n^p
g[n_, p_] := n*p
DirichletConvolve[f[n, p], g[n, p], n, 4]
First, we define two functions f and g. Then we compute their Dirichlet convolution.
The third argument in the Dirichlet convolution tells us that n is the function argument for which we want to do the convolution.
p on the other hand is a parameter that happens to exist in the functions but is not related to the convolution.
Changing the last line to
DirichletConvolve[f[n, p], g[n, p], p, 4]
means that we are using p as the variable for the convolution, whereas n now is some parameter.
Finally, the 4 says that we want to evaluate the resulting function at 4.
If you want to evaluate this function at the general position m you use
DirichletConvolve[f[n, p], g[n, p], n, m]
Mathematical Way
Let me write this in a mathematical way: We have two functions
$$ f \colon \mathbb{N} \times \mathbb{N} \longrightarrow \mathbb{N} \\ (n,p) \longmapsto n^p $$
and
$$ g \colon \mathbb{N} \times \mathbb{N} \longrightarrow \mathbb{N} \\ (n,p) \longmapsto n\cdot p $$
Now DirichletConvolve[f[n, p], g[n, p], n, m] evaluates
(f*g)(m, p) = \sum_{d \mid m} f(d, p) g \left(\frac{m}{d}, p\right)
$$ (f*g)(m, p) = \sum_{d \mid m} f(d, p) g \left(\frac{m}{d}, p\right) $$
whereas DirichletConvolve[f[n, p], g[n, p], p, m] evaluates
(f*g)(m, p) = \sum_{d \mid m} f(p, d) g \left(p, \frac{m}{d}\right)
It turns out that I can not render these formulas as LaTeX, because my reputation is too low :-(
$$ (f*g)(m, p) = \sum_{d \mid m} f(p, d) g \left(p, \frac{m}{d}\right) $$
Example from the Documentation
If the functions f and g do not have any parameters, this looks like:
f[n_] := n (* or any other function depending only on n *)
g[n_] := n
DirichletConvolve[f[n], g[n], n, m]
This example is equivalent to the one from the documentation
DirichletConvolve[n, n, n, m]
We convolve the identity map with itself and evaluate it at m.