= Solution
Let $C\in\operatorname{Neg}(K_X)$. The <arithmetic adjunction formula on a smooth surface> gives
$$
2p_a(C)-2=C^2+K_X\cdot C,\qquad p_a(C)\ge0.
$$
We first exclude $C^2\ge0$, following the PDF's hint and the finiteness proved in (a).
If $C^2=0$, adjunction and $K_X\cdot C<0$ force $p_a(C)=0$ and $K_X\cdot C=-2$. An integral projective curve of <arithmetic genus> zero is a <smooth rational curve>: normalization and the nonnegative singularity-length correction show both normalization genus and singularity correction vanish. Part (iii) makes $C$ <semiample>. Choose a basepoint-free $|mC|$. Over the infinite algebraically closed <field>, a general section avoids containing any of the finitely many curves of $\operatorname{Neg}(K_X)$ as a component. Its effective divisor $G$ then has $K_X\cdot G\ge0$, but $K_X\cdot G=mK_X\cdot C<0$, a contradiction.
If $C^2>0$, <Riemann–Roch theorem for algebraic surfaces> and <Serre duality> show $h^0(X,mC)$ is unbounded. Indeed $h^2(X,mC)=h^0(X,K_X-mC)=0$ for $m\gg0$, because the latter divisor has negative intersection with a fixed <ample> divisor. Hence
$$
h^0(X,mC)\ge\chi(X,\mathcal O_X)+\tfrac12(m^2C^2-mK_X\cdot C)\longrightarrow\infty.
$$
There is therefore some $m$ with $h^0(X,mC)>h^0(X,(m-1)C)$, giving a section whose divisor does not contain $C$. The canonical section of $mC$ does not vanish identically along any other curve. A general linear combination of these two sections consequently contains none of the finite set $\operatorname{Neg}(K_X)$, again contradicting its negative canonical intersection. Thus $C^2<0$.
Now put $a=-C^2\ge1$ and $b=-K_X\cdot C\ge1$. Adjunction gives
$$
-a-b=2p_a(C)-2\ge-2.
$$
It follows that $a=b=1$ and $p_a(C)=0$. Hence
$$
\boxed{C\cong\mathbb P^1,\qquad C^2=-1,\qquad K_X\cdot C=-1.}
$$
The general-section argument only avoids finitely many proper linear subspaces, so it works over any algebraically closed field, including positive characteristic.
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