The decreasing order of bond angles in BF$$_3$$, NH$$_3$$, PF$$_3$$ and I$$_3^-$$ is:

First, we recall the VSEPR (Valence Shell Electron Pair Repulsion) principle which states that electron pairs surrounding a central atom arrange themselves in space so that the repulsion between them is minimized. The order of repulsion is

$$\text{lone pair-lone pair} \; > \; \text{lone pair-bond pair} \; > \; \text{bond pair-bond pair}.$$

More repulsion between pairs forces the bonded atoms farther apart, thus producing a larger bond angle. Conversely, less repulsion gives a smaller bond angle.

Analyzing each species one by one

We have $$\text{BF}_3$$. Boron contributes three valence electrons and each fluorine donates one electron for the bond, giving a total of three bond pairs and, importantly, no lone pair on the central boron. According to VSEPR, $$3$$ bond pairs adopt a trigonal-planar geometry. The standard bond angle for a perfect trigonal plane is

$$120^\circ.$$

Next, we look at $$\text{NH}_3$$. Nitrogen has five valence electrons. It forms three $$\sigma$$ bonds with three hydrogen atoms (three bond pairs) and is left with one lone pair. Thus, there are $$4$$ regions of electron density (three bond pairs + one lone pair). VSEPR tells us that the electron pairs would choose a tetrahedral electron-pair arrangement, whose ideal bond angle is $$109.5^\circ$$. However, one of these positions is occupied by a lone pair, and we already know

$$\text{lone pair-bond pair repulsion} > \text{bond pair-bond pair repulsion}.$$

This extra repulsion squeezes the three $$\text{N-H}$$ bonds closer together, reducing the angle from the ideal $$109.5^\circ$$ to about

$$107^\circ.$$

Now consider $$\text{PF}_3$$. Phosphorus is in the same group as nitrogen but one period lower. It also has five valence electrons, so it similarly forms three $$\sigma$$ bonds with fluorine and retains one lone pair. Thus the qualitative geometry is again trigonal-pyramidal. Nevertheless, two additional factors decrease the bond angle relative to $$\text{NH}_3$$:

(i) The larger size of the central phosphorus nucleus spreads the bonding pairs slightly farther from each other.

(ii) Fluorine is highly electronegative, pulling bonding electron density toward itself. This diminishes the bond-pair electron density close to phosphorus, so $$\text{bond pair-bond pair}$$ repulsion becomes weaker. A weaker repulsion allows the angle to contract further.

Experimentally the $$\text{F-P-F}$$ bond angle is found to be close to

$$96^\circ.$$

Finally we examine $$\text{I}_3^-$$. The central iodine atom is surrounded by two bond pairs (the two terminal I atoms) and three lone pairs, giving a total of $$5$$ electron pairs. According to VSEPR, five electron pairs adopt a trigonal-bipyramidal electron-pair geometry. Of these, the three lone pairs preferentially occupy the equatorial positions (to minimize lone-pair-lone-pair repulsion), while the two bond pairs occupy the two axial positions. Axial positions in a trigonal-bipyramidal arrangement are located in a straight line, so the $$\text{I-I-I}$$ bond angle becomes

$$180^\circ,$$

i.e. the molecule is linear.

Putting the values in descending order

$$180^\circ \; (\text{I}_3^-) \; > \; 120^\circ \; (\text{BF}_3) \; > \; 107^\circ \; (\text{NH}_3) \; > \; 96^\circ \; (\text{PF}_3).$$

Writing the corresponding molecular formulas in the same sequence gives

$$\text{I}_3^- \; > \; \text{BF}_3 \; > \; \text{NH}_3 \; > \; \text{PF}_3.$$

This matches exactly with Option A in the question statement.

Hence, the correct answer is Option A.