Consider the general hypothetical reaction A(s)⇌2B(g)+3C(g)
If the partial pressure of C at equilibrium is doubled, then after the equilibrium is re-established, the partial pressure of B will be:
KP=P3C×P2B
Also, KP=P13C×P12B
P′C=2PC
⟹8P3C×P′2B=P3C×P2B
∴P′B=PB2√2=12√2 times the original value.