= Solution
There is no external horizontal <force>, so <momentum conservation> keeps the student–punt <centre of mass> fixed if the system starts at rest. Let the student's walking direction be positive. If the punt's <displacement> relative to the water is $X$, the student advances $X+L$, where $L=3\,\mathrm m$ is measured relative to the punt. Constancy of the <centre of mass> gives
$$
M_pX+m_s(X+L)=0,
\qquad
\boxed{X=-\frac{m_sL}{M_p+m_s}=-\frac{80\times3}{150+80}\,\mathrm m\simeq-1.04\,\mathrm m.}
$$
The punt moves $1.04\,\mathrm m$ opposite to the student's walking direction; the student's net <displacement> relative to the water is $3+X\simeq1.96\,\mathrm m$. These displacements are relative to the initial state, not distances travelled relative to the punt.
Under the same ideal assumptions, merely changing the walking speed or walking back more slowly cannot produce unlimited travel. Returning the student to the original place on the punt restores the original internal <mass> distribution and hence the punt's original <position>. A finite punt bounds the displacement available through internal rearrangement. Propulsion over large distances requires an external horizontal <impulse>, for example a push against the bed or <momentum> transferred to the water. Such an interaction changes the isolated, frictionless model rather than evading <centre of mass> conservation.
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