Let $f:X \to Y$ be a proper surjective morphism of projective surfaces such that there exists a curve $C \subset X$ for which $f_{X\backslash C}$ is an isomorphism and $f(C)$ is a set of points. Suppose $X$ is a closed subscheme of $\mathbb{P}^n$ for some integer $n$ and $C$ contracts to a rational singularity i.e., $f(C)$ are rational singularities on $Y$. Does there exist a closed immersion of $Y$ into $\mathbb{P}^n$ for the same integer $n$?
Let $X\subset \mathbb P^3$ be an arbitrary smooth projective surface of degree $d>2$ and assume that $X$ contains a line $\ell\simeq \mathbb P^1$ If $d=4$, assume in addition that $X$ is general among such surfaces. This implies that its Picard number is $2$. A simple adjunction computation shows that $\ell^2=2d<0$ ($l^2$ computed on $X$). This implies that $\ell\subset X$ is contractible, so we have a morphism $f:X\to Y$ as requested.
Claim. $Y$ cannot be embedded into $\mathbb P^3$. If $d>4$, then $Y$ cannot be embedded into any smooth $3$fold.
Proof:
 If $d=3$ (case abx): Since $X$ is rational, so is $Y$, so if it could be embedded to $\mathbb P^3$, its degree would have to be $1,2$ or $3$. Its easy to see using adjunction and that fact that $f$ is the blow up a smooth point that none of these is possible.
 If $d=4$ (case Olivier): The Kodaira dimension of $X$ is $0$, so if $Y$ were embeddable, it would have to have degree $4$. In this case adjunction does not help, because $0=K_X=f^*K_Y$. However, if $H'\subset Y$ is the hyperplane section of $Y\subset \mathbb P^3$, then $f^*H'\cdot\ell =0$ and since the Picard number of $X$ is $2$, we can compute that in this case $f^*H'=2mH+m\ell$ for some $m\in \mathbb N_+$. But then $H'^2=18m^2\neq 4$, so this leads to a contradiction and hence $Y$ cannot be embedded into $\mathbb P^3$.
 If $d>4$ (case new):This is actually the easiest. If $Y$ embeds into any smooth $3$fold, then it is Gorenstein and $K_Y$ is a Cartier divisor. Then we have $K_X=f^*K_Y+ a\ell$ for some $a\in \mathbb Z$. Using adjunction for $\ell$ and our earlier computation tells us that $$2d=\ell ^2=\frac {2}{a+1}$$ which means that $$2=(d2)\cdot(a+1),$$ that is, $d2$ divides $2$ so $d>4$ is out of the question. This completes the proof.

$\begingroup$ @Kovacs: Thank you for the very detailed and exhaustive answer. $\endgroup$ Jan 4 '14 at 14:30
No. Take a smooth cubic surface $S\subset \mathbb{P}^3$, and a line $E\subset S$. Then $E$ can be contracted to give a smooth Del Pezzo surface of degree 4 (a complete intersection of two quadrics in $\mathbb{P}^4$), which cannot be embedded in $\mathbb{P}^3$.
Let $X$ be a projective complex K3 surface whose NéronSeverigroup is generated by two classes $h$ and $l$ such that $h^2=4$, $hl=1$ and $l^2=2$ and $h$ is ample. One sees that such a surface exists using the surjectivity of the period map for K3 surfaces. General theorems about linear systems on K3 surfaces imply that $h$ is very ample and embeds $X$ in $\mathbb{P}^3$.
Since $l^2=2$, $l$ or $l$ is effective. Since $hl=1$, it is $l$ that is effective, and it has to be the class of a line in $\mathbb{P}^3$. This line is a $2$curve and may be contracted to a node. Let $f:X\to Y$ be this contraction map. Since $Y$ is a singular K3 surface, if it were possible to embed it in $\mathbb{P}^3$, the degree of the embedding would necessarily be $4$, but this is impossible as no line bundle on $Y$ has this degree.

$\begingroup$ @Benoist: Could you say why $Y$ has no line bundle of degree $4$? $\endgroup$ Jan 3 '14 at 19:03

$\begingroup$ @user43198: Its pullback to $X$ would be a degree $4$ line bundle on $X$ that is orthogonal to $l$ (because the line that is a section of $l$ is contracted by $f$). But line bundles on $X$ that are orthogonal to $l$ are proportional to $2h+l$ and $(2h+l)^2=18$. $\endgroup$ Jan 3 '14 at 21:45