solution concentrations
Since you did not tell us how much of this final solution you would like to make, let's imagine that you want to make a generic $V_f\ \ce{mL}$ of it.

Take $\text{A}$ as an example:

$$
n_A \ \text{in final solution} =  25000\ \ce{ng/mL} \times V_f\ \ce{mL}
$$

$$
V_{A}\ \text{stock to take} = \frac{n_A}{c_{A\ stock}} = \frac{25000\ \ce{ng/mL} \times V_f\ \ce{mL}}{1\ \ce{mg/mL} \times 10^6\ \ce{ng/mg}} \frac{1000\ce{\ \mu L}}{\ce{mL}}= \frac{25000}{10^3}\ V_f\ \ce{\mu L}
$$

So let's suppose that you want to make $10\ \text{mL}$ of such a solution containing $\text{A}$ at $25,000\ \text{ng/mL}$, then you need to measure out $25000 \times 10\ /\ 1000\ \ce{\mu L} = 250\ \ce{\mu L}$ of the $\text{A}$ stock and bring it up to the final volume with urine.

You will want to carry out these calculations for all your compounds, then add all the corresponding quantities to a calibrated vessel, e.g. a volumetric flask, then bring this mixture up to the final volume with your solvent, i.e. urine.