Step 1: List every phase separately, don't lump solids together.
A common mistake is assuming all solids form one phase. That's wrong unless they form a single homogeneous solid solution. Here $\mathrm{Pb_3O_4(s)}$ and $\mathrm{PbO(s)}$ are two different compounds with distinct crystal structures, so they're two separate solid phases. The gas, $\mathrm{O_2(g)}$, is always one phase. Total: $P=2\text{ (solids)}+1\text{ (gas)}=3$.
Step 2: Count species and subtract for the reaction constraint.
Species present: $\mathrm{Pb_3O_4}$, $\mathrm{PbO}$, $\mathrm{O_2}$, so 3 species, tied together by the single equilibrium $2\mathrm{Pb_3O_4}\rightleftharpoons6\mathrm{PbO}+\mathrm{O_2}$. The general rule is:
\[ C = (\text{species}) - (\text{independent reaction equilibria}) - (\text{extra restrictions}) \]
There is 1 reaction and no further restriction, so $C=3-1-0=2$.
Step 3: Plug into the phase rule and solve for $F$.
$F=C-P+2=2-3+2=1$.
Step 4: Interpret the answer physically as a check.
$F=1$ means this is univariant: exactly one intensive variable, say temperature, can be chosen freely, and everything else, including the equilibrium $\mathrm{O_2}$ partial pressure, is then fixed. This matches the known behavior of solid-solid-gas decomposition equilibria of this type.
Final Answer:
$P=3$, $C=2$, $F=1$, option (A).
\[\boxed{P=3,\ C=2,\ F=1}\]