Tone, I will sort something out for you, although lord only knows why other than extending the arcane knowledge of that peashooter of yours. There is a big mixture of terms in the above thread and not all are applied correctly.
However as has been stated by the mighty Bewsh, it is a calculation. Muzzle energy as above is calculated in simple terms as 0.5 x mass of object x (velocity of object)2[squared]. It's imperial measurement is ft lb (note not foot squared lb).
Pressure = Force/Area, and as Limulus says, Newtons Second Law applies. This in simple terms says Force = mass x acceleration (for those that know, I know this isn't correct, but getting into a conversation about rate of change of momentum is probably not going to get far - happy to have it though).
So, by substitution, Pressure = (mass x acceleration)/Area.
Now we need to make some assumptions to make this easier and let's pick the point in time as the bullet strikes, i.e. the surface area of a ballistic tip is very small. Now in essence dividing by a very small number gives you a very large pressure, so your calc, wasn't quite rght anyway - you needed to divide by (1/144 in your case), which is the same effect as multiplying by 144. Now pick the distance at which you wish to fire the bullet at the target and we can extrapolate an acceleration from that (and the calibre obviously).
I'm going to do this in modern (SI) units rather than imperial, as it comes out in the wash a bit easier.
Assume 243 (specially for you Tone). Assume bang on the legal limit for deer at 1700 ft lb or 2305 Joules. Assume 100gr bullet = (64.8mg x100) = 6.48g = 0.00648kg. Assume muzzle velocity derived from energy. Assume surface area of tip of bullet = 1mm square = 0.000001 m2.
the calculation then effectively derives 2305/0.000001 = 2,305,000,000 Pascals - so - a LOT then. In old money, 1 psi (pound per square inch) = 6,895 Pa, so 334,300 psi at full muzzle velocity.
HAPPY NOW?!