A beaker containing a liquid is placed inside a large closed jar....

The pressure at a depth $h$ near the bottom of the liquid is given by:

$P = P_{0} + \rho g h$

where $P_{0}$ is the pressure of the air inside the closed jar, $\rho$ is the density of the liquid, and $g$ is the acceleration due to gravity.

As the air within the jar is continuously pumped out, the pressure of the air $P_{0}$ above the liquid surface decreases.

Since $P_{0}$ decreases and the term $\rho g h$ remains constant (assuming the height of the liquid does not change significantly), the total pressure $P$ near the bottom of the liquid will decrease.