I have read your question several times to figure out what your problem is.
Below, I will try to give you a simple explanation without any fancy mathematics or complex equations.
Now, suppose we dissolve the weak acid "A" in water. The acid will react with water and form hydrogen ions and the corresponding base "B" of the weak acid. We further suppose that the pH of this solution is 3 at equilibrium.
We have the following equilibrium:
$$\ce{A + H2O <--> H3O+ + B}$$
The concentrations are supposed to be low, so we can state an equilibrium constant:
$K = [\ce{H3O+}]*[\text B]/[\text A]$
Now, we add a strong acid to this solution. What will happen?
The pH of the solution will drop. Suppose it will drop to pH = 2. We will get an immediate response from the equilibrium reaction of the weak acid system. Since the amount of hydrogen ions have increased in the solution by the addition of the strong acid, the equilibrium will be pushed to the left.
Pushed to the left? What does it mean?
Well, the new situation implies that the weak acid cannot give off the same amount of hydrogen ions as it did at pH = 3. pH is now 2 and we have a completely new situation.
Now I will target on what I think is your main problem.
When you add a strong acid to the weak acid solution, the pH will drop.
The equilibrium of the weak acid will then reset into a new equilibrium determined by the actual pH and the equilibrium constant of the system.
At pH = 3, hydrogen ions were formed from the weak acid. These (or other) hydrogen ions will be forced to react with the corresponding base B, giving more A at pH = 2.
Thus, the net effect is that the hydrogen ion concentration related to the strong acid will be in principle unchanged.