P is true if and only if A and B are independent if and only if $P(A \cap B) = P(A) \cdot P(B)$. If A and B are mutually exclusive, then $P(A \cap B) = 0$ (or $A \cap B \neq \empty$ depending on what kind of probability you are doing), thus Q is equivalent to the statement $P(A \cap B) \neq 0$.
Thus neither statement is true as stated: if event A is impossible, then trivially $P(A \cap B) = 0 \cdot P(B) = 0$ so $A, B$ are independent events, however $A$ and $B$ are mutually exclusive because $P(A \cap B) \leq P(A) = 0$. This could happen if, for instance, A is the probability of a coin flip creates an elephant on your table (impossible) and B is the probability that the coin comes up heads.
In the other direction Q does not imply P because, for instance, the probability of a 6-sided die coming up an even number could be A and the probability of the die coming up a 2 could be B. Clearly A and B are not independent as $P(A \cap B) = P(B) = 1/6$ which is not $P(A)\cdot P(B) = 1/12$.
However if we assume that neither A nor B has probability zero, then P necessarily implies Q: if $P(A \cap B) = P(A) \cdot P(B)$ and both $P(A)$ and $P(B)$ are nonzero, then the right hand side is nonzero, thus the left hand side must be nonzero as well. Thus A and B cannot be mutually exclusive.