Solution

Since the material of the bags is inextensible, the perimeter of their cross-section does not change during deformation. Initially, the perimeter of one bag is $C = 2\pi r$.

After the slab is placed on them, the bags are flattened to a thickness of $h \ll r$. In this flattened state, the cross-section of the bag can be approximated as a rectangle with width $w$ and height $h$. The perimeter of this cross-section is $2w + 2h$. Using the condition $h \ll r$, we can neglect $2h$, so the perimeter is approximately $2w$.
Equating the initial and final perimeters:
$$2w \approx 2\pi r \implies w = \pi r$$

Now we can write down the initial and final volumes for one bag:
Initial volume: $V_1 = \pi r^2 L$
Final volume: $V_2 = w \cdot h \cdot L = \pi r h L$

The slab of mass $m$ rests on two bags. The total horizontal contact area of the slab with both bags is:
$$S = 2 \cdot w \cdot L = 2\pi r L$$

In a state of mechanical equilibrium, the force of gravity acting on the slab is balanced by the difference between the force from the gas pressure inside the bags ($P_2$) and the external atmospheric pressure ($P_0$) acting on this contact area:
$$mg = (P_2 - P_0)S$$
From this, we express the final pressure in the bags:
$$P_2 = P_0 + \frac{mg}{2\pi r L}$$

According to the problem statement, the gas temperature does not change ($T = \text{const}$). Applying Boyle's Law for the initial and final states of the gas inside one bag:
$$P_1 V_1 = P_2 V_2$$

Substituting the expressions for the volumes:
$$P_1 (\pi r^2 L) = P_2 (\pi r h L)$$
$$P_1 \cdot r = P_2 \cdot h \implies P_1 = P_2 \frac{h}{r}$$

Substituting the expression for $P_2$ into this equation and expanding the brackets:
$$P_1 = \left( P_0 + \frac{mg}{2\pi r L} \right) \frac{h}{r} = P_0 \frac{h}{r} + \frac{mgh}{2\pi r^2 L}$$