Let $S$ be a algebraic surface (so a $2$-dimensional, proper $k$-scheme) and $\mathcal{L}$ a semi ample and globally generated invertible sheaf on $S$.
Since it is globally generated it provides a morphism $g: S \to \mathbb{P}^n$ for $n$ big enough such that $\mathcal{L}= g^*(\mathcal{O}_{\mathbb{P}^n}(1))$.
Let $E$ a irreducible curve on $S$ (so a $1$-dimensional, proper subscheme of $S$) such that $\mathcal{L} \vert _E \cong \mathcal{O}_E$ holds.
Why does it imply that already $g(E)= \{p\}$. Therefore $E$ is mapped to a point?
Why $g(E)$ can't be also a irreducible curve?
I suppose that we can use the condition $\mathcal{L} \vert _E \cong \mathcal{O}_E$ in some way to show that $g(E)$ can't be a curve, but how?
This is essentially projection formula (one can prove this without it, but morally it is the same). If $f:X\to Y$ is a morphism, $L$ a line bundle on $Y$, $C\subset X$ is a curve, then projection formula will say $(C\cdot f^*L)=(f_*C\cdot L)$. So, in your case if $f(E)$ is a curve, one has $f_*E=\deg(f:E\to f(E)) f(E)$ and thus since $\mathcal{O}_{\mathbb{P}^n}(1)$ is ample, $(f_*E\cdot \mathcal{O}_{\mathbb{P}^n}(1))>0$, which gives $(E\cdot L)>0$ where $L=f^*(\mathcal{O}_{\mathbb{P}^n}(1))$. But, your hypothesis implies $(E\cdot L)=0$. Thus $f(E)$ can not be a curve and so must be a point.