We need to find the incorrect statement.

Check option A: $\mathrm{PH_3}$ vs $\mathrm{NH_3}$ proton affinity.

• Proton affinity depends on the availability of the lone pair.
• In $\mathrm{NH_3}$, the lone pair on nitrogen is more available than in $\mathrm{PH_3}$ for stable bond formation with $\mathrm{H^+}$.
• Also, basicity order among group 15 hydrides is such that $\mathrm{NH_3}$ has higher proton affinity than $\mathrm{PH_3}$.

Therefore, A is correct.

Check option B: $\mathrm{SO_2}$ can act as oxidizing agent, but not as reducing agent.

• In $\mathrm{SO_2}$, sulfur is in oxidation state $+4$.
• Since sulfur can be oxidized from $+4$ to $+6$, $\mathrm{SO_2}$ can act as a reducing agent. Example: $\mathrm{SO_2 + Br_2 + 2H_2O \rightarrow H_2SO_4 + 2HBr}$ Here $\mathrm{SO_2}$ is oxidized, so it behaves as a reducing agent.
• Since sulfur can also be reduced from $+4$ to lower oxidation states (for example to $0$ or $-2$ in suitable reactions), $\mathrm{SO_2}$ can also act as an oxidizing agent.

Hence the statement "can act as an oxidizing agent, but not as a reducing agent" is false.

Therefore, B is incorrect.

Check option C: $\mathrm{NO_2}$ can dimerise easily.

• $\mathrm{NO_2}$ has one unpaired electron.
• It dimerises to form $\mathrm{N_2O_4}$: $2\mathrm{NO_2} \rightleftharpoons \mathrm{N_2O_4}$

Therefore, C is correct.

Check option D: $\mathrm{PF_3}$ exists but $\mathrm{NF_5}$ does not.

• $\mathrm{PF_3}$ is a known stable compound.
• Nitrogen cannot expand its octet because it has no vacant $d$-orbitals in the valence shell and is too small to accommodate five fluorine atoms.
• Hence $\mathrm{NF_5}$ does not exist.

Therefore, D is correct.

Conclusion:

The only incorrect statement is: $\boxed{\text{B}}$