The reducing strength of a substance is generally associated with its ability to donate electrons to other substances. In the context of hydrides of Group 15 elements, such as $\mathrm{SbH}_3$, $\mathrm{NH}_3$, $\mathrm{BiH}_3$, and $\mathrm{PH}_3$, the reducing strength can be considered based on their chemical reactivity, stability, and tendency to donate electrons.

In general, the stability of hydrides decreases as we move down the group in the periodic table, which means that heavier hydrides are less stable and hence act as stronger reducing agents. This trend is primarily due to the increase in the size of the central atom as we move down the group, leading to weaker bonds between the central atom and hydrogen in the hydride. As the bond strength decreases, the hydrides can more readily release hydrogen in the form of protons (H⁺) or hydride ions (H⁻), making them stronger reducing agents.

The order of reducing strength for Group 15 hydrides is typically as follows:

$$\mathrm{BiH}_3 > \mathrm{SbH}_3 > \mathrm{PH}_3 > \mathrm{NH}_3$$

Among the options given, $\mathrm{BiH}_3$ (bismuthine) is the strongest reducing agent. It is the least stable of the listed hydrides, due to the large size and low electronegativity of bismuth compared to antimony, phosphorus, and nitrogen. This makes $\mathrm{BiH}_3$ more prone to releasing electrons and acting as a reducing agent.

In contrast, $\mathrm{NH}_3$ (ammonia) is the most stable among the given hydrides and, therefore, the weakest reducing agent. Its bond with hydrogen is comparatively stronger, and nitrogen's higher electronegativity makes the molecule more resistant to donating electrons.

Thus, the correct answer is Option C, $\mathrm{BiH}_3$.