Elliptic curves $E$ and $E'$ over a finite field $K$ are $K$-isogenous if and only if the orders of $E(K)$ and $E'(K)$ coincide. However, it may happen that the groups $E(K)$ and $E'(K)$ have the same order (and even isomorphic) but $E$ and $E'$ are not isomorphic over $K$. Even worse, there exist such a $K$ and non-isomorphic over $K$ elliptic curves $E$ and $E'$ such that if $\bar{K}$ is an algebraic closure of $K$ then the Galois modules $E(\bar{K})$ and $E'(\bar{K})$ are isomorphic. In particular, if $L$ is an arbitrary finite field containing $K$ then the groups $E(L)$ and $E'(L)$ are isomorphic. (Of course, $E(K)$ and $E'(K)$ are the subgroups of Galois invariants in $E(\bar{K})$ and $E'(\bar{K})$ respectively.) See arXiv:0711.1615 [math.AG].
An explicit description of all groups that can be realized as $E(K)$ (for a given $K$) was done by Misha Tsfasman (In: Theory of numbers and its applications, Tbilisi, 1985, 286--287; see also Sect. 3.3.15 of the book Algebraic geometric codes: basic notions by Tsfasman, Vladut and Nogin, AMS 2007).
See also papers of René Schoof (J. Combinatorial Th. A 46 (1987), 183--211), Felipe Voloch (Bull. SMF 116 (1988), 455--458) and Sergey Rybakov (Centr. Eur. J. Math. 8 (2010), 282--288).
I don't think your map is injective. Here is an attempt at a recipe for constructing a counterexample.
The ingredients are a $C_1$-field $F$ of characteristic zero and a smooth projective curve $X_0$ over $F$ having non-trivial Brauer group. For a concrete example take $F=\mathbf{C}(t)$ and $X_0$ to be an elliptic curve with non-trivial Brauer group; a calculation of such a curve can be found in O. Wittenberg, Transcendental Brauer–Manin obstruction on a pencil of elliptic curves, PDF file here.
Edit: What's below is unnecessary. $X_0$ is already a counterexample, since every finite extension of $F$ has trivial Brauer group.
Given these ingredients, let $k$ be the field $F((t))$, let $\mathcal{X}$ be a smooth proper curve over $F[[t]]$ whose special fibre is $X_0$, and let $X$ be the generic fibre of $\mathcal{X}$.
There is a natural injective map $\mathrm{Br}(\mathcal{X}) \to \mathrm{Br}(X)$. Let $\alpha$ lie in the image of this map, and let $P$ be a closed point of $X$. If $R$ is the integral closure of $F[[t]]$ in $k(P)$, then $P$ extends to an $R$-point of $\mathcal{X}$, and $\alpha(P) \in \mathrm{Br}(k)$ lies in the image of $\mathrm{Br}(R)$, which is trivial ($R$ is a Henselian local ring whose residue field is $C_1$). Therefore $\alpha(P)=0$. This holds for all closed points $P$, so $\alpha$ lies in the kernel of your map.
It remains to show that $\mathrm{Br}(\mathcal{X})$ is non-trivial. For any $n$ coprime to the characteristic of $F$, proper base change gives $\mathrm{H}^2(\mathcal{X},\mu_n) \cong \mathrm{H}^2(X_0, \mu_n)$.
Then the Kummer sequence shows that $\mathrm{Br}(\mathcal{X})[n] \to \mathrm{Br}(X_0)[n]$ is surjective. In particular, $\mathrm{Br}(\mathcal{X})$ is non-trivial whenever $\mathrm{Br}(X_0)$ is non-trivial.
Remark: if you don't insist that $X$ is a curve, then things are much easier: take a surface over a finite field having non-trivial Brauer group.
Best Answer
The construction holds for any base field $k$. But if $k$ is not perfect, you get a projective curve which is normal but not necessarily smooth. For example, if $k$ has characteristic $p>2$ and $t\in k$ is a not a $p$-th power in $k$, consider the function field $K=k(x,y)$ defined by the relation $y^2=x^P-t$. The curve you get is not smooth, and there is no projective smooth curve over $k$ whose function field is $K$. Note that smooth curves are always normal, and the converse is true if $k$ is perfect.
Pick any transcendental element $x\in K$. Then $K$ is finite over $k(x)$. Let $A$ be the integral closure of $k[x]$ in $K$ and let $B$ be the integral closure of $k[1/x]$ in $K$. Then the localizations $A_x$ and $B_{1/x}$ are both equal to the integral closure of $k[x, 1/x]$ in $K$. Therefore we can glue the affine curves ${\rm Spec} A$ and ${\rm Spec} B$ along ${\rm Spec} A_x$ and get a curve $C$. By constuction $C$ is normal and integral, with field of functions $K$, and there is finite morphism $C\to \mathbb P^1={\rm Spec} k[x] \cup {\rm Spec} k[1/x]$ (obtained by glueing ${\rm Spec} A\to {\rm Spec} k[x]$ and ${\rm Spec} B\to {\rm Spec} (k[1/x])$). Hence $C$ is its projective, and it is the projective normal curve associated to $K$.
As a bonus, one also has a nice correspondance betweeen finite morphisms of curves and extensions of function fields of one variable. If $f : C\to D$ is a finite morphism of projective normal integral curves over $k$, then it induces a finite extension $k(D)\to k(C)$. One can show that this establises a anti-equivalence from the category of integral normal projective curves over $k$ (where morphisms are finite morphisms of $k$-curves) to the category of function fields of one variable over $k$ (where morphisms are morphisms of $k$-extensions).