The equations of motion of any matter, as well as Einstein's equation for gravitation are encoded in the Einstein-Hilbert Lagrangian :
$$
S=\int d^4x \sqrt{-g}\left[\frac{1}{16\pi G}R+\mathcal{L}_{\mathcal{M}} \right]
$$
See here for a bit more info.
The matter content of the theory (i.e., the particle you talk about for instance) will be contained in the Lagrangian $\mathcal{L}_{\mathcal{M}}$. For a massive particle for example (which trajectory is parametrised by $\tilde{x}^\mu(\lambda)$), the action would be :
$$
S_{partic}=-m\int d\lambda \sqrt{\frac{d\tilde{x}^\mu}{d\lambda}\frac{d\tilde{x}^\nu}{d\lambda}g_{\mu\nu}(\tilde{x}(\lambda))}
$$
Let us consider the simplest case in which the particle is affinely parametrised. This simplifies the action, and we can consider instead :
$$
S_{partic}=-\frac{m}{2}\int d\lambda \frac{d\tilde{x}^\mu}{d\lambda}\frac{d\tilde{x}^\nu}{d\lambda}g_{\mu\nu}(\tilde{x}(\lambda))
$$
For a single particle, it is not very convenient to use the field theory notation, so we will not extract $\mathcal{L}_{\mathcal{M}}$. Instead, the full action of the system becomes :
$$
S=\frac{1}{16\pi G}\int d^4x \sqrt{-g}R+S_{partic}
$$
The full action now depends on both the metric field $g_{\mu\nu}(x)$, as well as the particle trajectory $\tilde{x}^\mu(\lambda)$. To obtain the equations of motion, we should vary both w.r.t. to $g_{\mu\nu}$ and $\tilde{x}^\mu$. Doing the latter variation, the ricci scalar $R$ will not contribute as it does not depend on $\tilde{x}^\mu$.
So we obtain the equation of motion for a particle moving in a metric space $g_{\mu\nu}$, which is simply the geodesic equation (here I write it assuming $\lambda$ is an affine parameter) :
$$
\frac{d^2 \tilde{x}^\mu}{d^2\lambda}+\Gamma^\mu_{\rho\nu}\frac{d\tilde{x}^\rho}{d\lambda}\frac{d\tilde{x}^\nu}{d\lambda}=0
$$
Where $\Gamma^\mu_{\nu\rho}$ are the Christoffel symbols for the metric.
Now, we must do the variation w.r.t. $g_{\mu\nu}(x)$, the metric. Under this variation, we have :
$$
\delta(\sqrt{-g}R)=\sqrt{-g}(R_{\mu\nu}-\frac{R}{2}g_{\mu\nu})\delta g^{\mu\nu}
$$
It remains to compute the variation of $S_{partic}$. We will have to write $g_{\mu\nu}(\tilde{x}(\lambda))=\int d^4x g_{\mu\nu}(x)\delta^4(\tilde{x}-x)$. This is because we want to make a variation w.r.t. $g_{\mu\nu}(x)$, so we should express it as a function of $x$.
To be clearer, this is like when we say $\frac{\delta f(a)}{\delta f(x)}=\delta(x-a)$. The precise steps are $f(a)=\int dx f(x) \delta(x-a)$. Then $\delta f(a)=\int dx \delta f(x) \delta(x-a)\equiv \int \frac{\delta f(a)}{\delta f(x)}\delta f(x)$, which gives the aforementioned result.
Anyway, after rewriting the metric, and using as $\delta g_{\mu\nu}=-g_{\mu\rho}g_{\lambda\nu}\delta g^{\rho\nu}$ we find :
\begin{align}
\delta S_{partic}&=\frac{m}{2}\int d\lambda\frac{d\tilde{x}^\mu}{d\lambda}\frac{d\tilde{x}^\nu}{d\lambda}\int d^4x \delta g_{\mu\nu}\delta^4(x-\tilde{x})\\
&=-\frac{m}{2}\int d\lambda \int d^4x \left(\frac{d\tilde{x}_\rho}{d\lambda}\frac{d\tilde{x}_\nu}{d\lambda}\right)\delta g^{\nu\rho} \delta^4(x-\tilde{x})
\end{align}
Putting everything together, we obtain the following equations :
$$
R_{\mu\nu}-\frac{R}{2}g_{\mu\nu}=16\pi G\left(\frac{m}{2}\int d\lambda \frac{d\tilde{x}_\nu}{d\lambda}\frac{d\tilde{x}_\mu}{d\lambda} \delta^4\left(x-\tilde{x}(\lambda)\right)\right)
$$
On the RHS, this is simply the Stress-energy tensor of a single particle. All in all, we end up with two differential equations which are coupled. Solving these coupled equations will give you both the particle trajectory, and the metric of spacetime, which will include the backreaction of the particle on the spacetime as it traverses it.
However, these coupled equations are very difficult to solve. What one does in practice is assume the backreation of the particle on the spacetime is small. This will have for effect to decouple the system of equation. Then, what you would do is first solve the Einstein equations to find the metric, and THEN solve the particle geodesic equations to find the particle trajectory. In that approach, you neglect how the particle will deform the spacetime as it moves.
