As said in the comments to your question, the quantum numbers assigned to particles and the resonances in order to fit experimental data, will give the allowed reactions. It is simpler to think in terms of the possible Feynman diagrams for the specific decay one is looking at (similar to the one in the title) for example:

Particles can decay via the strong interaction, and if such a decay pathway is available to a particle, it decays very quickly - on the order of $10^{-23}$ seconds. An example is the decay $Δ_0 → p^+ + π^-. Note that the delta baryon Δ0 has the same quark makeup as the neutron, but its mass is much larger. Its mass is sufficient for this decay to be energetically favorable.
Then apply all the conservation laws at the vertices , by looking at the assignments in the PDG table for the specific resonance or particle. This automaticaly takes care of parity and charge conjugation.
Charge conjugation Parity reversal, depends on the quantum numbers assigned in the table, Time reversal :
Time reversal (T): replacing t by -t. This reverses time derivatives like momentum and angular momentum.
CPT invariance ties them into one relation:
Examples in nature can be cited for the violation of each of these symmetries individually. It was thought for a time that CP (parity transformation plus charge conjugation) would always leave a system invariant, but the notable example of the neutral kaons has shown a slight violation of CP symmetry.
So the answer to your title is that the decay happens because it is not violating any conservation law. The individual symmetry violations are taken into account by using the assigned quantum numbers .
CPT invariance automatically happens in all particle interactions we have studied and modeled into the standard model. A violation of CPT would hit at the basic premises of the present theory, (as has been argued Lorenz invariance), for which there is no experimental evidence.