First lets clarify one thing: Light mesons, wich are the entities envolved in this decay, have all one thing in common: their orbital angular momentum is l=0. They are, however, grouped in the pseudo-scalar mesons (with s=0 for the pions for exemple) and the vector-mesons (with s=1 for the rho and omega for example).
The parity of a meson state is the product of the parity of its constituents by the parity of it's orbital wave function like this: P(q)xP(q*)x(-1)^l where q* is the anti-quark and the term (-1)^l is the parity of the orbital wave function. Check Modern Particle Physics (Mark Thompson) on page 229!
Thus, since this is a strong decay, the parity must be conserved from the initial state to the final state:
P(omega)x(-1)^l = P(rho)xP(pion)x(-1)^l
(here we can consider the center of mass referential of the two resulting products and refer to the orbital angular momentum there without loss of generality because we arealy know tha l=0) Now, since P(q)=1 and P(q*)=-1:
P(q)xP(q*)x(-1)^l = P(q)xP(q*)xP(q)xP(q*)x(-1)^l
1x(-1)x(-1)^0 = 1x(-1)x1x(-1)x(-1)^l
-1 = 1*(-1)^l
What can be concluded in order to conserver parity is the following: the orbital wave function of the final state must present an odd value for l (like l= 1, 3 etc) meaning that s waves (l=0), d waves (l=2) etc are excluded and only p waves (l=1), f waves (l=3) etc can be observed.
Furthermore this must be consistent with the conservation of angular momentum:
The initial state (Ji) has Ji=1 ( since J=s+l=1+0)
The final state (Jf) has the following sum : Jf= s1+s2+l where s1 and s2 are the spins of the rho and pi mesons. Acording to the rules of addition of angular momenta we first add two of them: s1+s2 = 1+0 = 1 and then we add the third giving Jf=1+l. We can then conclude that l must either be 0, 1 or 2 since this is the only way we can get Jf to equal Ji.
(remember that to add angular momentum: |l1-l2|,|l1-l2|+1,|l1+l2|+2,.... stopping only when we have he value |l1+l2| and so, the values 1 and 2 will spam the sbspace with J=1 as well!). Thus we can clearly see that in this case, to respect both parity and angular momentum ocnservation the final state l must really be l=1.
Hope this gives a clear sight on what is happening in this deacay!