Hybridization and excitation of electrons, how is it possible?
We don't really consider the excitation of electrons. If it appears like this, then this is probably due to an incorrect understanding of what an orbital actually is: The attempt to describe each single electron by its own wave function which is independent of all the other electrons (which it is not). 

An orbital is defined as single electron wave function $\varphi(\vec r_i)$ where $\vec r_i$ are the coordinates of electron $i$. We then try to approximate the many electron wave function as an (anti-symmetrized) product of orbitals

\begin{equation}
   \Psi(\vec r_1, \vec r_2, \dots \vec r_N) \approx \prod_{i=1}^N \varphi(\vec r_i)
\end{equation}

But this is strictly speaking not true, because the electrons are correlated with each other, due to the Coulomb repulsion $V\propto \frac{1}{|r_i-r_j|}$. The product ansatz does not work for the kind of differential equation we need to solve. The Hartree-Fock method tries to do so anyway, and gives therefore wrong (or at least only approximate) result.

This means the whole concept of orbitals is an approximation, which may give a qualitative understanding, but are failing when trying to get quantitative results.

One approach to correct this approximation is expanding the many electron wave function in some basis and solve the resulting eigenvalue problem. This basis is not unique, but one possible choice is to use the Hartree-Fock determinant and the excited configurations you can generate from it. But these excited configurations are more of an abstract mathematical thing and have limited physical meaning, because they are based on the above approximation. The real physics/chemistry is in the solutions of such methods. And those solution cannot be describe by the above ansatz for $\Psi$.