Does $AD^{L(\mathbb{R})}$ directly implies projective determinacy? At least it certainely implies $PD$'s consistency.

8$\begingroup$ Yes, of course it does. All projective sets are obviously in $L (\mathbb R) $ so they have winning strategies (in $L (\mathbb R) $ and therefore in $V $; note that any possible set of moves for either player is coded by a real). $\endgroup$– Andrés E. CaicedoOct 1 '17 at 12:11
Suppose $M$ is an inner model (of $\mathsf{ZF}$) with the same reals as $V$, and let $A\subseteq \mathbb R$ be a set of reals in $M$. Suppose further that $A$ is determined in $M$. Under these assumptions, $A$ is also determined in $V$. The point is that since winning strategies are coded by reals, and any possible run of the game for $A$ is coded by a real, then any run of the game in $V$ is already in $M$.
Similarly, since $M$ and $V$ share the same reals, they agree on what sets are firstorder definable over the reals, that is, they agree on what sets of reals are projective. It follows that if projective determinacy holds in $M$ then it holds in $V$. In fact, much more is true in the sense that $M$ and $V$ agree on many more definitions over the reals than just firstorder (for instance, $L(\mathbb R)^M=L(\mathbb R)$).
It follows immediately from these two observations (taking $M=L(\mathbb R)$) that if determinacy holds in $L(\mathbb R)$, then in $V$ all sets of reals in $L(\mathbb R)$ are determined. :) That is, $\mathsf{AD}^{L(\mathbb R)}$ implies $L(\mathbb R)$determinacy. In particular, projective determinacy holds in $V$.
Naturally, $V$ may have more sets of reals than $M$, some of which may not be determined. Also, in general, there may be notions of definability that do not coincide in $M$ and $V$ (for instance, by forcing we may easily change what sets of reals are in $\mathsf{HOD}$). It may be that some definition $\phi$ gives in $M$ a set $A$ that is determined in $M$. As long as we do not change the reals, that set $A$ will be determined in $V$, but running the same definition $\phi$ in $V$ may result in a different set $\hat A$ that may very well fail to be determined.