Suppose that $p \colon P \to X$ is a projective bundle. Is $p$ an open map ?

For any $X,Y$ the projection $X \times Y \to Y$ is an open map. Since being open is a local property you know this holds also for any fiber bundle and in particular any projective bundle. 


Isn't any fiber bundle $p: E \to B$ projection an open map? If $e \in E$ belongs to an open $U$, then there is a trivializing neighborhood $V_e$ of $p(e)$ so that $p^{1}(V_e) \to V_e$ is isomorphic to a projection map $V_e \times F \to V_e$, which is certainly open. Hence $p(U \cap p^{1}(V_e))$ is open. Since direct images preserve unions, we see that $$p(U) = p(\bigcup_{e \in U} U \cap p^{1}(V_e)) = \bigcup_{e \in U} p(U \cap p^{1}(V_e))$$ is open. 

