First of all we have to determine the functional cover of the set of functional dependencies as given to determine all the non-trivial functional dependencies those can be generated from the given set of functional dependencies.
F={M->O,NO->P,P->L,L->MN}
Now F+={M->O,NO->P,P->L,L->MN,L->M,L->N,NO->L,NO->MN,MN->P,OTHER TRIVIAL DEPENDENCIES}
We have two decompositions
R1={L,M,N,P}
Let F1 be the set of functional dependencies which will projected on R1 from F+.
So F1={P->L,L->MN,L->M,L->N,MN->P,OTHER TRIVIAL DEPENDENCIES}
We have R2={M,O}
So F2={M->O,OTHER TRIVIAL DEPENDENCIES}
Now F’=$F1\cup F2={P->L,L->MN,L->M,L->N,MN->P,M->O,OTHER TRIVIAL DEPENDENCIES}$
F’+={P->L,L->MN,L->M,L->N,MN->P,M->O,P->MN,P->M,P->N,OTHER TRIVIAL DEPENDENCIES}
now if we compare F’+ and F+ we can see that NO->P and NO->L dependencies are not present in F’+ which are present in F+.
So this decomposition is not dependency preserving.
But this is lossless join decomposition as common attribute M is the candidate key in relation R2.
Now R2 is BCNF as it is a two attribute relation.
For R1 we have to determine the candidate keys from F1.
And we can easily determine the candidate keys are P,L,MN.
and all the dependencies in the F1 preserve the property of BCNF i.e LHS of all the FDs should be a superkey.
So R1 is also BCNF.