There are some number of scattered photons that still have their original
incident wavelength (lambda). This is because they collided with an electron
that was still strongly bound to its parent atom and so the photon didn't
pass any of its energy as it would to a free electron.
I slightly disagree with the explanation given in the video
(linked below in your comment).
The left peak in all the sub-figures of your image
(with the unchanged wavelength $\lambda$ and with angle $\theta$)
are not photons scattered by an electron, but photons scattered
by a nucleus.
The shift of wavelength for scattering by an electron is given by
Compton's formula:
$$\lambda'-\lambda=\frac{h}{m_ec}(1-\cos\theta)
=2.4\cdot 10^{-12}\text{ m}\cdot(1-\cos\theta)$$
The atomic binding energy of the electron (even for a stronger bound
inner electron) is neglectable. See the last section of hte answer.
And the shift of wavelength for scattering by a nucleus is much
smaller due to the bigger mass $m_C$ of the carbon nucleus:
$$\lambda'-\lambda=\frac{h}{m_Cc}(1-\cos\theta)
=0.00011\cdot 10^{-12}\text{ m}\cdot(1-\cos\theta)$$
These results are for Compton's experiment where the photon target
was Graphite. Do we know if this experiment was tested on any target
other than Graphite in which the electrons were more or less strongly
bound to the parent atom and see how the results differ?
Compton's experiments is done with high-energy photons (X-rays or gamma
rays), i.e. with photon energies higher than $1$ keV. This is much higher
than the atomic binding energy of the electron. E.g. the binding energy
of an electron in graphite is only a few eV (even for the electrons
in the more tightly bound inner shells). Therefore the atomic binding
energy of the electrons can be safely neglected when compared to the
energy the X-ray photon.
As far as I know, Compton's experiment has been repeated also with
metals and non-metals. And it gives essentially the same results
as with graphite. This was to be expected by the reasoning above.
