"Geometry" is complicated in quantum mechanics, because we have probability distributions rather than precise geometrical positions as we would have in classical mechanics. If we make the Born-Oppenheimer approximation, we can remove some of that complexity and actually talk about the geometry of nuclear coordinates (separated from the quantum mechanically behaving electronic coordinates) get pictures like the one below, in which we have the square of the complex modulus of the wavefunction (which by Born's rule is the probability of finding the system at a certain location) plotted as a function of a nuclear coordinate (or geometry):
We can say that the "geometry" of the $v=1$ state is the geometry (or set of nuclear coordinates) at which the probability (or square of the complex modulus of the wavefunction) is greatest, so you would look for the maximum value of the purple curve corresponding to $v=1$ in the above figure, for example. In the above figure, the potential energy curve/surface looks quite symmetric, as in the case of a harmonic oscillator, so it looks like we have two geometries with equally maximal probability, but the more anharmonic the potential, the less likely that that this would be the case.
Therefore, to find the most-probable geometry for a $v=1$ state, you can determine the wavefunctions of the system, and from them you can determine the probability functions as plotted in the above figure.
To get the wavefunctions, you can use the procedure suggested in my answer to your previous question: Algorithm for finding the v=1 state of an H2O molecule