If the particle moves with group wave, what $\lambda$ in De Broglie equation should we use? according to De Broglie equation
\begin{gather} p=\frac{h}{\lambda} \end{gather}
and knowing also that a particle moves with the group velocity not the phase velocity, indicates that has a range of $\lambda$'s inside it superpositioned with each other.
then in the De Broglie's equation What $\lambda$ should we use? if the particle when moving has many of $\lambda$'s inside it's wave, then what $\lambda$ should we use in this eqaution?
 A: In general that equation is only for particles that have a specific $\lambda$, or even a specific momentum. In general a particle need not be in a state which has those properties well defined. So at any given moment you may not be able to assign a wavelength to a particle in any reasonable way.
Looking back, the idea that a particle of matter had a definite wavelength was more of a historical thing that was "on the right track" while the quantum mechanics of today was being developed, which allows for superpositions of different wavelengths. Today we still use the notion sometimes for heuristics but not for exact descriptions except in very special circumstances where the wave is strongly centered on just one wavelength.
A: Well, then you have to operate on momentum and wavelength ranges, like :
$$ \Delta p \Delta \lambda = h \tag 1$$
And this a lot reassembles uncertainty principle (lower bound), except for the $\hbar$ and $1/2$ factor.
Interesting thing. If you'll rewrite (1) as
$$ m\Delta v \times \frac {\Delta v}{\Delta \nu} = h \tag 2,$$
now multiply each side by $1/2$, we'll get :
$$ \frac {m \Delta v^2}{2} \times \frac {1}{\Delta \nu} =\frac h2 \tag 3$$
First term is kinetic energy uncertainty, and second - wave period uncertainty, so :
$$ \Delta E_k \times \Delta T = \frac h2 \tag 4,$$
which now is similar to uncertainty principle for energy-time. Except that here is particle kinetic energy specifically and wave oscillation period uncertainty and of course we have lost $\hbar$ factor here also.
But what I wanted to say overall, that if particle momentum and wavelength ranges must somehow be included into De Broglie relationship, then in the end - you will loose De Broglie equation meaning and you will be better operating with Heisenberg uncertainty principle instead.
