So here is my physics problem. I have built a "new and improved" (read heavier) drop hammer. I am curious how many pounds per square inch will be created upon impact on my die.

The weight is 50 Pounds.
The distance is 5 feet.
The surface area of the die is 1.25 square inches.

Using the Internet I am able to get it down to Kinetic Energy (337.84 J), but don't know if that can be converted into pounds per square inch.

Calculating impact pressure is a complicated physical problem that depends upon the properties and deformation of the object impacted, not just the simple laws of physics that were used to calculate the impact energy (338 joules). That's why you couldn't find a simple calculation on the Internet.

For the average impact force during the very brief process of stopping the hammer, you could write:

Impact Force x Impact (stopping) Distance = Impact Energy

If we assume a stopping distance of 1 millimeter (0.001 m) corresponding to the deformation of a typical coin, then in SI units:

Impact Force = 338 Newton-meters (joules) / 0.001 meters = 338 kN

Converting to English units and calculating the pressure for your specific case:

Pressure = 338 kN / (4.45 N/lbf) / 1.25 sq. in. = about 60 ksi (60,000 psi).

This is certainly in the correct range for the plastic deformation of metals at room temperature ("cold working"), but it could easily be off by a factor of 2 or more depending on the assumption about stopping distance and the details of what happens upon impact.

Five feet is not high enough for my drop hammer to strike my one ounce gold with a simple drop. Next test, I will “throw” it down and see if that makes a difference. Who knows, maybe the third test will be a hot strike.