I didn't say it contributes to reaction rate
I'm going to just ignore this.
I am no Niels Bohr but I would deduce that the net/effective pressure of the system would increase.
Yes. The 'total pressure' of the system would increase.
Partial pressure is the contribution of a particular species towards the total pressure. If n, v and T are fixed, then this contribution remains constant, regardless of how much other gases you add in (provided n doesn't change due to any reactions, of course). We use partial pressure because it is a good measure of the rate of collision between gaseous species, and since equilibrium is all about equal forwards and backwards rates, it is the best quantity which we can use to describe gas equilibria.
You can relate addition of inert gas to the gas equation in two ways (assuming fixed container and constant temperature):
1. pp.V=n.R.T, where pp=partial pressure. In this case, adding more inert gas does not add more reagent gas, its partial pressure does not change, and the rate of reaction is not altered.
2. P.V=N.R.T, where P=total pressure, and N=total number of moles of all gaseous species (assuming they're all ideal gases). Adding more inert gas increases the total number of moles N, and thus the total pressure must increase. However, the total pressure doesn't tell us anything about the rate of reaction.
The total pressure is the sum of all partial pressures. This comes from the linear relationship between p and n in pV=nRT.
 RT \\<br />PV & = n_1 RT + n_2 RT + n_3 RT + \cdots \\<br />PV & = pp_1 V + pp_2 V + pp_3 V + \cdots \\<br />PV & = (pp_1 + pp_2 + pp_3 + \cdots)V \\<br />P & = pp_1 + pp_2 + pp_3 + \cdots<br />\end{align*}<br />)