No, they aren't.

Pounds per inch, pounds per foot, pounds per whatever are running loads or a type of distributed loads.

Example:
A load of 870 pounds is uniformly distributed along a 10 foot beam.  The running load is 870/10 = 87 pounds per foot = 87 lb/ft.

If the load is not distributed uniformly we can still describe the magnitude of the load as a running load.

Torque and moment are usually expressed as pounds-inch or pounds-foot in the American Customary System of Measure.  A moment is a force acting through an arm of some distance, say a weight on the end of a beam sticking out of a wall.  A torque produces a twisting action on a component.

Example (this might require a sketch if you aren't accustomed to thinking in "pictures"):

Consider an L shaped object with 10 inch long arms.  Hold the L horizontal, with both arms parallel to the floor.  Add a 10 pound weight at the end of one arm so gravity tries to pull it toward the floor, and grab the end of the other leg in your hand, resisting the weight.

The moment applied by the weight to the leg you are holding is 10 inchesX10 pounds = 100 lb in or in lb, whichever order you prefer.  Your hand holding the leg is reacting that 100 in lb moment with a 100 in lb torque.

This is one part of a fundamental principal from mechanics in which the sum of forces and the sum of moments must balance.  And while I was writing that, it occurred to me that a screwdriver with a perpendicular handle might have been a better example, one that about anyone reading here has experienced first hand.

On to pressure:
Pressure is a load distributed over an area, pressure = load / area = pounds per inches squared, pounds per feet squared, pounds per area. I.e., another type of distributed load.

Example:
The 870 pound load above is applied to a plate of 20 inches X 5 inches.  The area of the plate is 100 square inches, or 100 inches^2 expressed mathematically.
Then the pressure on the plate is 870 pounds / 100 inches^2 = 87 lb/in^2.

Now, it's time for a bonus question -

What is the running load on each edge of the 20X5 inch plate in the example above?

Back to the trigger question -

What are you trying to determine?  Trigger pull is measured in terms of the force required to release the sear.

A pressure exerted by a finger pad on a trigger could be calculated, but first you need the trigger pull weight and then the contact area where the finger and trigger interface.  Not too interesting practically.

The force to trip the sear can easily range from 2 ounces on a benchrest gun, to 10 and 12 pounds on factory handguns.

The other issue you'll need to consider is travel required.  That ranges from barely perceptible, if at all, to way too long.  Some triggers include an overtravel stop to limit trigger movement after break; that ranges from zero to unrestricted within the limits of the basic assembly.

For the pressure you calculated, that will be no issue on an aluminum or steel trigger.  It's approaching the limit for hard, reinforced plastics such as the aluminum filled epoxy we use as liquid shim material.  Your larger problem with a plastic part might be compliance, basically lost motion due to the soft spring rate of the material.  Also note that many if not most triggers have arched, not flat, faces, so the initial contact is  point, not a line or area.

Mostly I think you're looking at a trivial third order part of your design problem and you need to focus on just about everything else.