Question: Because, physically, if the volume is constant and the pressure increases, won't it mean that there is an increase in that internal energy?

If not, how else should we perceive it?

Answer: 
It’s called an isochoric (constant volume) heat transfer process. For an ideal gas the ratio of pressure to temperature is constant and the change in internal energy is $C\_{V}\Delta T$.

Hope this helps
Let's illustrate this for an ideal gas. (In general, we should be considering thermodynamic potentials.) Using the ideal gas law $pV=nRT$, we can put this in differential form by noting that
$$d(nRT) = d(pV)\Longrightarrow nR dT = Vdp+pdV,$$
where we have assumed that the system is closed ($n=$ constant). Now, during a constant volume process, $pdV=0$, and so we get
$$nRdT=Vdp.$$
Finally, for an ideal gas, the internal energy is given by
$$U=\frac{f}{2}nRT,$$
where $f$ is the number of degrees of freedom (per particle) of the gas, in which case
$$dU = \frac{f}{2}nRdT.$$
Putting these together, we get
$$Vdp=\frac{2}{f}dU,$$
and so while $Vdp$ itself is not a change in energy of the system, it is *proportional* to the change in internal energy during a constant-volume process. Finally, since the volume is constant, no work is done on the system, and so the total change in energy is due only to heating.