Propositional formula in DNF can be decided in polynomial time?
The conversion from CNF to DNF can come at an exponential cost. For example $(a_1 \lor b_1) \land \cdots \land (a_n \lor b_n)$ expands to $2^n$ many terms. As you comment, for DNF satisfiability is easy - it is falsifiability which is hard. If the problem is trivial, you don't input it to a SAT solver, and that's why SAT solvers accept CNFs instead of DNFs.

If you believe that P is different from NP, then this implies that there is no polynomial time way to convert CNF satisfiability to DNF satisfiability, since the former is NP-complete while the latter is in P.