Can we take the eigenvalue we obtained after momentum operator (of quantum mechanics) operates on the state vector of a system as the momentum of the system?
No. Mathematically multiplying your state vector by an operator is not how you determine the outcome of a measurement. First, that would assume you can get a deterministic value, which we know isn't true of measurements of quantum systems. Second, if your system is (most likely) in a superposition of momentum states, then this operation will not even give you a single value for momentum, as your state is not an eigenstate of momentum.
The only way this works is if your state was an eigenstate of momentum before the momentum measurement. Then by coincidence operating on your state vector with the momentum operator will give you the momentum of the system as that eigenvalue. However, states like these are not physical, so this case cannot be realized.
Yes, if it's in a pure momentum eigenstate, which is the only way you would obtain a single eigenvalue of the momentum operator. There are complications due to the fact that you can't really have a system in a pure momentum state (though you can get arbitrarily close). But conceptually, that's what it means to have a single eigenvalue.
Note: applying the momentum operator does not change the state of the system. The way to do that, i.e., get a system into a pure eigenstate, is to perform a measurement, which collapses the wavefunction.