Let $P(X,Y,Z)$ denote a homogeneous polynomial in $\mathbb{C}[X,Y,Z]$ such that $X_P = \{(u : v : w) \in \mathbb{C}\mathbb{P}^2 \mid P(u,v,w) = 0\}$ defines a smooth complex projective curve in $\mathbb{C}\mathbb{P}^2$. It inherits a Riemannian metric from the Fubini-Study metric on $\mathbb{C}\mathbb{P}^2$.
Is there an explicit formula or algorithm that gives the area of $X_P$ as a real surface in terms of the polynomial $P(X,Y,Z)$?
A much more general result is given by Mumford, in Projective varieties I, Theorem 5.22: the volume of any $r$-dimensional smooth projective variety in $\Bbb{P}^N$ (the area in your case) is its degree times the volume of a linear subspace $\mathbb{P}^r$. The latter is easily calculable, depending on how you normalize the Fubini-Study metric.
The Fubini-Study 2-form $\alpha$ on complex projective space is a calibrating form in the sense that it evaluates to 1 on every complex direction, and is less than 1 on all other real 2-dimensional directions. Therefore integration of $\alpha$ over any complex curve will give precisely the area of the curve, whereas for any embedded Riemann surface, the integral will provide a lower bound for the area. This technique is applied very often in Riemannian geometry, following Marcel Berger and Mikhael Gromov after him.
Since the integral is a homology invariant by Stokes, the area is proportional to the degree of the algebraic curve.
Here's a sketch of proof for the curve case. The volume form associated to the Fubini–Study metric is the restriction of the Fubini–Study symplectic form $\omega_{\rm FS}$ (which is a 2-form on $\mathbb{CP}^2)$. Therefore the area of a curve $C$ is given by the integral of $\omega_{\rm FS}$ over $C$, which is just the evaluation of the cohomology class of $\omega_{\rm FS}$ on the homology class of $C$, $$ {\rm Area}(C) = \int_C \omega_{\rm FS} = \langle [\omega_{\rm FS}], [C] \rangle. $$ Since $[C] = d[L]$, where $L$ is a complex line and $d$ is the degree, we get that the area is just $d$ times the area of a line.
The same argument should work more generally, recovering the result quoted in abx's answer. (Just take the appropriate power of $\omega_{\rm FS}$.
