If $p,q$ are two projections in a $C^*$-algebra such that $p\leq q$, we have $p$ is subordinate to $q$. But the converse may not be true.
Question1.Can we add some additional conditions such that the following conclusion holds: $p$ is subordinate to $q$, then $p\leq q$.
Yes. Your condition is equivalent to the algebra being commutative, as long as the algebra is generated by projections (which every von Neumann algebra is).
Assuming the condition that $q\preceq p$ implies $q\leq p$, if the algebra is generated by projections, given any unitary $u$ and any projection $p$, the projection $upu^*$ is equivalent to $p$, so $upu^*\leq p$. As you can do the same with $u^*$, you get that $upu^*=p$ for all unitaries and all projections, so the algebra is commutative.
For general C$^*$-algebras (i.e., non-von Neumann) the question is kind of meaningless. For instance take $B$ to be any projectionless C$^*$-algebra, like $C_0(\mathbb R)$ if you want commutative, or $C_0(\mathbb R)\otimes K(\ell^2(\mathbb N))$ if you want non-commutative. Then your property holds (as there are no projections where it could fail) but you don't get anything out of it. Same is true with projectionless unital C$^*$-algebras, like $C_r^*(\mathbb F_2)$.