Why is the Paschen Law linear at large pd values?

First, it's only approximately linear.

The general idea is that an electron needs a certain amount of energy to ionize a neutral. During ionization, the neutral releases an electron, which along with the initial electron, is accelerated by the electric field. Both electrons can ionize new neutrals once they have enough energy, bringing the number up to four, which proceed to ionize more atoms, etc. This cascade is called an avalanche breakdown.

Enough new ions need to be generated so that at least one strikes the cathode and liberates a new electron (not all ions hitting the cathode will create an electron via secondary electron emission). This electron can then enter the gap and restart the process. This is brief rough description of Townsend discharge.

That being said, electrons need to have a long enough mean free path so that they can gain enough energy to ionize a neutral before they undergo a collision. If the pressure is too high (really density), the electrons dink around with many collisions and never gain enough energy to ionize. Thus you need higher electric fields to accelerate the electrons to the necessary energies to ionize in a shorter distance. Since mean free path is inversely proportional to pressure of the gas, an increase in pressure corresponds an proportional increase in electric field needed to maintain the proper acceleration.

Increasing the gap between the electrodes without changing the pressure or voltage reduces the electric field, thus reducing acceleration and energy imparted to the electrons. So if the d is increased, the voltage needs to be increased by a corresponding amount to maintain the electric field strength.

From a math viewpoint, the slope of the Paschen curve at larger pd is:

$$ slope approx frac{B}{ln{(A times pd)}} - frac{B}{ln{(A times pd)}^2}$$

where A and B are constants specific to the gas. The above slope is roughly constant over reasonably large spans of pd.