Let us first consider a capacitor that is charged and not connected to a battery or other electric power source.
I think you need to take into account that the Electric field in the capacitor is reduced by inserting the dielectric and also the voltage drops between the plates by inserting the capacitor. The dielectric is pulled in to the cap as the capacitor loses electrical potential energy as its voltage is decrease.
The dielectric in the capacity gets polarized so that the surface of the dielectric facing the positively charged plate becomes negatively charged and the surface of the dielectric facing the negatively charged plate becomes positvely charged. The charges on each side of the dielectric are less than the charges on the plates. Overall the electric field inside the capacitor (inside the dielectric) is reduced because the effective total charge on each side of it is reduced.
The electric field inside the capacitor $E'$ is reduced by
$E'={1 \over k} E~~~~~$ or $~~~~~E'={1 \over \epsilon_r} E$
The reduction in electric field comes with a reduction in the potential between the two plates so the energy stored in the capacitor is reduced as the same ammount of charge is stored in it, but the voltage is now lower - so that if it is discharged the power term $VI$ would be smaller as $V$ is lower.
If the battery is still attached then as the dielectric is inserted the capacitance is increased and the capacitor charges up more. The charge is multiplied by dielectric constant, $k$, and then the battery supplies the extra charge and so it loses energy.
As the battery is attached the electric field inside the dielectric is the same as the original electic field in between the plates without the dielectric, but now the charge is higher on the plates and there is again charge induced on the surfaces of the two dielectrics. I think the upper of the two diagrams in the question is correct because the electric field in the gap is the same whether with and without the dielectric inserted.
There will be addition charge density on the plates in regions II and III, but these will be balanced by the extra opposite charges on the surface of the dielectric in II and III so that the upper diagram is correct.
Best Answer
A dielectric slab will be attracted towards any source of electric field, because it is essentially a collection of dipoles. Dipoles in fields align with the fields. If the field varies in space, the dipole is drawn to the local maximum of the field. An intuitive, everyday example of this is the way iron filings are drawn to a magnet.
In the case of a constant voltage capacitor, the battery supplies the extra energy. We can see this most simply in the case that we replace the battery with a very large capacitor of capacitance $C$ connected in parallel with our original capacitor with capacitance $c_0 \ll C$. Originally, each capacitor is at a voltage $V_0$, and the large capacitor has charge $Q_0=CV_0$ while the small capacitor has charge $q_0=c_0V_0$. As we insert the dielectric, $c_0$ changes by $\Delta c$. The capacitors then settle in the new condition of equal voltage
$$\frac{Q_0-\Delta q}{C} = \frac{q_0 + \Delta q}{c_0 + \Delta c}$$
Solving for $\Delta q$ using the fact that $c_0 +\Delta c \ll C$ gives
$$\Delta q = Q_0 \frac{c_0 + \Delta c}{C} - q_0$$
From this it is fairly easy to show that the energy change in the original capacitor is
$$\Delta U_c=\frac{1}{2}\Delta c V_0^2$$
What about the change in the large capacitor? There we have
$$\Delta U_C = \frac{1}{2C}((Q-\Delta q)^2 - Q^2) \approx -\frac{\Delta q\, Q}{C}$$
Substituting gives
$$\Delta U_C=-\Delta c V_0^2$$
So overall the system ends with a lower energy, and the slab should be attracted.