What is the source of the unregulated voltage that feeds the LM7805? You maybe able to power the opamp from that directly without using the 7808. Remember that the Typical modern opamp has a Power Supply Rejection Ratio of > 90db, so the 8V does not need to be regulated. If fact the opamp's power pin could be anywhere from 8 to 30V, and the circuit would work just fine...
The PT100's resistance vs temperature curve is quite linear (that is its biggest virtue), so you do not need a look-up table, just a little math inside the PIC after you get the 10bit AD reading. The function which turns an AD value into temperature is:
T = (AD*300)/1024 = (AD*75)/256 where T is in degC and AD is the right-justified 10bit AD reading.
Note that the divide by 256 is just a right-shift by 8 bits. Any C compiler is smart enough to divide by 256 by using a shift instruction instead of linking in the math library...
To multiply by 75 without having to use the math library is:
75*AD = AD + 2*AD + 8*AD + 64*AD = (1+2+8+64)*AD
2*AD is the AD value left-shifted by 1 bit,
8*AD is the AD value left-shifted by 3 bits,
64*AD is the AD value left-shifted by 6 bits.
Most decent C compilers will implement a "multiply by a constant" 75 just that way.
The data sheet for the PT100 you bought will show how to hook it up. I'm guessing that two of the wires are connected to one end of the nominal 100 Ohm resistance and the other wire goes to the other end. If you use an Ohmmeter to test it, you will find what looks like a dead short between two of the wires. Just short those two together before putting it in the circuit.
The feedback resistor for the opamp should be 9000.0 Ω (that is much harder to source than 90.9K
). As Eric said, use a trimpot so that the total resistance is 9K (8.2K in series with a 1K trimpot would work). Use that to replace the parallel combination of R5//R6.