The number of bridged oxygen atoms present in compound B formed from the following reactions is
$$PbNO_{3_2} \xrightarrow{673 K} A + PbO + O_2$$
$$A \xrightarrow{Dimerise} B$$

When lead nitrate is heated at 673 K:

$$2Pb(NO_3)_2 \xrightarrow{673K} 4NO_2 + 2PbO + O_2$$

So compound A is $$NO_2$$ (nitrogen dioxide).

When $$NO_2$$ dimerizes:

$$2NO_2 \xrightarrow{Dimerise} N_2O_4$$

Compound B is $$N_2O_4$$ (dinitrogen tetroxide).

The structure of $$N_2O_4$$ consists of two $$NO_2$$ units connected by an N-N bond:

$$O_2N - NO_2$$

Each nitrogen atom is bonded to two oxygen atoms and to the other nitrogen atom. There are no bridged (bridging) oxygen atoms in this structure. All oxygen atoms are terminal, bonded to only one nitrogen atom each.

The number of bridged oxygen atoms = 0

Hence, the correct answer is Option A.