Is a morphism whose all fibers are $\mathbf{P}^n$ a projective bundle? Suppose $X\to Y$ is a morphism between varieties with all fibers isomorphic to $\mathbf{P}^n$, is $X$ a projective bundle over $Y$, i.e. $X=\mathbf{P}(E)$ for some vector bundle over $Y$?
 A: No, as Jason Starr explains. (I had misread the question before. Let me just give the OP a positive result in this direction, not completely unrelated to the question.)
Let $X\to Y$ be a proper flat (locally finitely presented) morphism of schemes such that all geometric fibres are isomorphic to projective $n$-space. (This implies that $X\to Y$ is smooth.)  
Let $I:= Isom_Y(X,\mathbb P^n_Y)\to Y$ be the Isom-scheme of $X$ with projective $n$-space over $Y$.   As the fibres of $X\to Y$ are projective spaces, for all $y$ in $Y$, the vector space $\mathrm{H}^1(X_y, T_{X_y}) =0$. Here $T_{X_y}$ is the tangent bundle of $X_y$ (over the residue field of $y$ in $Y$). Now, note that if $H^1$ of the tangent bundle vanishes, the Isom-schemes of $X$ are unobstructed. (Here we use flatness of $X\to Y$.)
Thus, $I\to Y$ is formally smooth. As it is locally finitely presented (by assumption), we conclude that $I\to Y$ is smooth. By assumption on $X\to Y$ it is surjective (all fibres are isomorphic to $\mathbb P^n$).
Now, note that $X_I = \mathbb P^n_I$ (because $I(I)$ contains the identity morphism). In other words, $X$ is isomorphic to $\mathbb P^n_Y$ over $I$. 
Since $I$ is smooth surjective, we conclude that $X$ is isomorphic to $\mathbb P^n_Y$, locally for the smooth topology on $Y$. Therefore, $X$ is isomorphic to $\mathbb P^n_Y$ over $Y$, locally for the etale topology on $Y$ (as by a theorem of Grothendieck smooth surjective morphisms have sections, locally for the etale topology). 
A: No that is not true, and I am confident somebody has written the following example previously on MO.  Let $\mathbb{A}^2$ have coordinates $s$ and $t$.  Let $Y\subset \mathbb{A}^2$ be the singular plane curve with defining equation $$f(s,t) = t^2-s^2(1-s) = 0.$$  Let $u$ be a coordinate on $\mathbb{A}^1$.  The normalization of $Y$ is $$\nu:\mathbb{A}^1 \to Y, \ \nu(u) = (1-u^2,(1-u^2)u).$$  Let $V\subset \mathbb{A}^1$ be the open complement of the closed point $p$ where $u=1$.  Let $X$ be $V\times \mathbb{P}^n$.  Denote by $$\text{pr}_V:V \times \mathbb{P}^n\to V,$$ the usual projection.  Then one counterexample is $\pi = \nu\circ \text{pr}_V$.  
Edit.  One reference for this problem is Exercise II.7.10 of Hartshorne's "Algebraic Geometry".  If $\pi$ is a proper, flat morphism $\pi$ as in the question, then Ariyan's answer is perfectly correct and gives étale local triviality. If, further, $Y$ is Noetherian and regular (the hypothesis in Exercise II.7.10), then generic triviality of $\pi$ (i.e., the OP's hypothesis at the generic point of $Y$) implies Zariski local triviality of $\pi$. There should be an easy direct argument for this, but this certainly follows from the known cases of the Grothendieck-Serre conjecture (there is a direct argument after all, see the hint in that exercise).  Finally, Exercise II.7.10 implies that $X$ is isomorphic to $\mathbb{P}(E)$.
