How can I find the symmetry labels of atomic orbitals in a molecule?
First, you have to know the geometry of your compound. The complex $\ce{[PtCl4]^2-}$, for example, is square planar.
The next step is to determine the point group of the compound from this geometry by identifying the symmetry elements that it posesses (a procedure on how to do this can be found here). The point group of the complex $\ce{[PtCl4]^2-}$, for example, is $D_\mathrm{4h}$.
Now you consult the character table of the point group you determined in the previous step. The character tables for many common point groups can be found here. The character table of the point group $D_\mathrm{4h}$ is shown below and can also be found here.


$$\begin{array}{c|cccccccccc|cc} \hline
D_\mathrm{4h} &amp; E &amp; 2C_4 &amp; C_2 &amp; 2C_2' &amp; 2C_2'' &amp; i &amp; 2S_4 &amp; \sigma_\mathrm{h} &amp; 2\sigma_\mathrm{v} &amp; 2\sigma_\mathrm{d} &amp; &amp; \\ \hline
\mathrm{A_{1g}} &amp; 1 &amp; 1 &amp; 1 &amp; 1 &amp; 1 &amp; 1 &amp; 1 &amp; 1 &amp; 1 &amp; 1 &amp; &amp; x^2+y^2,z^2 \\
\mathrm{A_{2g}} &amp; 1 &amp; 1 &amp; 1 &amp; -1 &amp; -1 &amp; 1 &amp; 1 &amp; 1 &amp; -1 &amp; -1 &amp; R_z &amp; \\
\mathrm{B_{1g}} &amp; 1 &amp; -1 &amp; 1 &amp; 1 &amp; -1 &amp; 1 &amp; -1 &amp; 1 &amp; 1 &amp; -1 &amp; &amp; x^2-y^2 \\
\mathrm{B_{2g}} &amp; 1 &amp; -1 &amp; 1 &amp; -1 &amp; 1 &amp; 1 &amp; -1 &amp; 1 &amp; -1 &amp; 1 &amp; &amp; xy \\
\mathrm{E_g} &amp; 2 &amp; 0 &amp; -2 &amp; 0 &amp; 0 &amp; 2 &amp; 0 &amp; -2 &amp; 0 &amp; 0 &amp; (R_x,R_y) &amp; (xz,yz) \\
\mathrm{A_{1u}} &amp; 1 &amp; 1 &amp; 1 &amp; 1 &amp; 1 &amp; -1 &amp; -1 &amp; -1 &amp; -1 &amp; -1 &amp; &amp; \\
\mathrm{A_{2u}} &amp; 1 &amp; 1 &amp; 1 &amp; -1 &amp; -1 &amp; -1 &amp; -1 &amp; -1 &amp; 1 &amp; 1 &amp; z &amp; \\
\mathrm{B_{1u}} &amp; 1 &amp; -1 &amp; 1 &amp; 1 &amp; -1 &amp; -1 &amp; 1 &amp; -1 &amp; -1 &amp; 1 &amp; &amp; \\
\mathrm{B_{2u}} &amp; 1 &amp; -1 &amp; 1 &amp; -1 &amp; 1 &amp; -1 &amp; 1 &amp; -1 &amp; 1 &amp; -1 &amp; &amp; \\
\mathrm{E_u} &amp; 2 &amp; 0 &amp; -2 &amp; 0 &amp; 0 &amp; -2 &amp; 0 &amp; 2 &amp; 0 &amp; 0 &amp; (x,y) &amp; \\ \hline
\end{array}$$


Finally you need to determine the symmetry labels of the p- and d-orbitals within the point group by identifying the irreducible representations (the labels on the far left of the character table) which match the transformational properties of the basis vectors ($\color{red}{x}$, $\color{red}{y}$, $\color{red}{z}$) and some of their quadratic combinations ($\color{blue}{xy}$, $\color{blue}{xz}$, $\color{blue}{yz}$, $\color{blue}{z^2}$, $\color{blue}{x^2 - y^2}$). These can be found in the two columns on the right columns. Again this will be shown for the example compound $\ce{[PtCl4]^2-}$ where the relevant functions are highlighted.


$$\require{bbox} \begin{array}{c|cccccccccc|cc} \hline
D_\mathrm{4h} &amp; E &amp; 2C_4 &amp; C_2 &amp; 2C_2' &amp; 2C_2'' &amp; i &amp; 2S_4 &amp; \sigma_\mathrm{h} &amp; 2\sigma_\mathrm{v} &amp; 2\sigma_\mathrm{d} &amp; &amp; \\ \hline
\mathrm{A_{1g}} &amp; 1 &amp; 1 &amp; 1 &amp; 1 &amp; 1 &amp; 1 &amp; 1 &amp; 1 &amp; 1 &amp; 1 &amp; &amp; x^2+y^2,\bbox[cyan]{z^2} \\
\mathrm{A_{2g}} &amp; 1 &amp; 1 &amp; 1 &amp; -1 &amp; -1 &amp; 1 &amp; 1 &amp; 1 &amp; -1 &amp; -1 &amp; R_z &amp; \\
\mathrm{B_{1g}} &amp; 1 &amp; -1 &amp; 1 &amp; 1 &amp; -1 &amp; 1 &amp; -1 &amp; 1 &amp; 1 &amp; -1 &amp; &amp; \bbox[cyan]{x^2-y^2} \\
\mathrm{B_{2g}} &amp; 1 &amp; -1 &amp; 1 &amp; -1 &amp; 1 &amp; 1 &amp; -1 &amp; 1 &amp; -1 &amp; 1 &amp; &amp; \bbox[cyan]{xy} \\
\mathrm{E_g} &amp; 2 &amp; 0 &amp; -2 &amp; 0 &amp; 0 &amp; 2 &amp; 0 &amp; -2 &amp; 0 &amp; 0 &amp; (R_x,R_y) &amp; \bbox[cyan]{(xz,yz)} \\
\mathrm{A_{1u}} &amp; 1 &amp; 1 &amp; 1 &amp; 1 &amp; 1 &amp; -1 &amp; -1 &amp; -1 &amp; -1 &amp; -1 &amp; &amp; \\
\mathrm{A_{2u}} &amp; 1 &amp; 1 &amp; 1 &amp; -1 &amp; -1 &amp; -1 &amp; -1 &amp; -1 &amp; 1 &amp; 1 &amp; \bbox[pink]{z} &amp; \\
\mathrm{B_{1u}} &amp; 1 &amp; -1 &amp; 1 &amp; 1 &amp; -1 &amp; -1 &amp; 1 &amp; -1 &amp; -1 &amp; 1 &amp; &amp; \\
\mathrm{B_{2u}} &amp; 1 &amp; -1 &amp; 1 &amp; -1 &amp; 1 &amp; -1 &amp; 1 &amp; -1 &amp; 1 &amp; -1 &amp; &amp; \\
\mathrm{E_u} &amp; 2 &amp; 0 &amp; -2 &amp; 0 &amp; 0 &amp; -2 &amp; 0 &amp; 2 &amp; 0 &amp; 0 &amp; \bbox[pink]{(x,y)} &amp; \\ \hline
\end{array}$$

Since the three p-orbitals point directly along the $x$, $y$, and $z$ axes, they show the same transformational properties as the basis vectors, and consequently get the same symmetry labels, i.e.

$$\begin{align}
\left.\begin{array}{c}
\mathrm{p}_x \\ \mathrm{p}_y \end{array}\right\} &amp;\rightarrow \mathrm{E_u} \\
\mathrm{p}_z &amp;\rightarrow \mathrm{A_{2u}} \ .
\end{align}$$

The five d-orbitals show the same transformational properties as their quadratic function counterparts and thus they get the symmetry labels, i.e.

$$\begin{align}
\mathrm{d}_{xy} &amp;\rightarrow \mathrm{B_{2g}} \\
\left.\begin{array}{c}
\mathrm{d}_{xz} \\ \mathrm{d}_{yz} \end{array}\right\} &amp;\rightarrow \mathrm{E_g} \\
\mathrm{d}_{z^2} &amp;\rightarrow \mathrm{A_{1g}} \\
\mathrm{d}_{x^2 - y^2} &amp;\rightarrow \mathrm{B_{1g}} \ .
\end{align}$$

Another worked example can be found in this answer of mine.