Acid/base nature of any reagent is a relative term. For instance, water $(\mathrm{p}K_\mathrm{w} = 14)$ also acts as an acid or base based on which reagent it is reacting with. Basically, any reagent with $\mathrm{p}K_\mathrm{a} \lt 7$, water acts as an base (e.g., acetic acid; $\mathrm{p}K_\mathrm{a} = 4.75$):
$$\ce{CH3COOH + H2O <=> CH3COO- + H3O+}$$
Then again, if the conjugate acid of any reagent has $\mathrm{p}K_\mathrm{a} \gt 7$, water acts with that reagent as an acid (e.g., the conjugate acid of ammonia with $\mathrm{p}K_\mathrm{a} = 9.25$):
$$\ce{NH3 + H2O <=> NH4+ + OH-}$$
Now, let's look at the given reactions:
(a) $\ce{(C2H5)2NH}$ + $\ce{H2PtCl6}$:
This is the though one among all four choices. The name of $\ce{H2PtCl6}$ is hexachloroplatinic acid, and thus, it is safe to assume that this acts like an acid in water meaning $\mathrm{p}K_\mathrm{a}$ of $\ce{H2PtCl6}$ is $ \lt 7$. Ammonia with water acts as a acid $(\mathrm{p}K_\mathrm{a} \gt 7)$ and hence the amine in this case must acts as a base.
(b) $\ce{CH3NH2}$ + $\ce{H2O}$:
$\ce{CH3NH2}$ acts like ammonia with water because $\mathrm{p}K_\mathrm{a}$ of its conjugate acid is about $10.5$. Thus, $\ce{CH3NH2}$ acts like a base here.
(c) $\ce{((Me)2CH)2NH + n-C4H9Li}$:
$\ce{n-C4H9Li}$ is a strong base with $\mathrm{p}K_\mathrm{a} \approx 50$ for conjugate acid of $\ce{n-C4H9-}$, $\ce{n-C4H10}$. Therefore, the amine acts here as an acid, giving away its lone proton.
(d) $\ce{(C2H5)3N + BF3}$:
Here, $\ce{BF3}$ is a Lewis acid, $\ce{B}$ in which does not fulfil the octet. Thus, it is craving for a electron pair, $\ce{N}$ in $\ce{(C2H5)3N}$ would gladly donate. Therefore, amine acts as a base here.
Therefore, no doubt, the choice C is the answer.
