maybe You don't need a blend pot, just do it like a reverb circuit:

R1 and R2 are mix resistors. you might as well use a pot instead of R1 and R2 to figure the ratio, but I'd think it'd be static once you figured it out.
the reason I don't think you need a blend pot is the combination of P2 and "V" pot will determine the levels of the two.
I'd make the R3+P1 series resistance be high, at least 1M, and make R1 and R2 be higher as well, like (if its 50/50 mix) 1M/1M, or if end up being a 1:4 mix, make R1 be 1M and R2 be 4M. This is so V1b's load will be high. I think the load of V1b is (R3+P1||R1) (is that correct?)
also, I'd try moving the .02/4.7M pair between P1 and the 68K.
and I don't see a pentode anywhere. If V1b is a pentode, then like tubeswell says, make load the of V1b high, so that R3+P1||R1>=1.5M or >=2M
edit: that 56K grid leak on the next stage, V3a might need a tweak when adding mix resistors (or a blend pot). i'm not sure what the original schematic looked like, if there was series resistor between the .001 coupling cap and the grid of V3a, then the ratio of the voltage divider should be taken into account as the next mix resistors will form voltage dividers with that 56K, you might want to play with that valve as well.