In the given reaction
$ RCOOH\xrightarrow{AgN{O_3}}\xrightarrow{B{r_2},{\Delta}}[P]$
[P] will be:
(a) R-Br
(b) RCOOR
(c) RCOOAg
(d) $$RC{H_2}Br$$

Hint: The silver nitrate is a chemical compound which has the ionic bonding present between the silver cation and nitrate anion. It has a colourless appearance in a solid state and it is odourless also. It has the tendency to be soluble in water. The structure of the silver nitrate is orthorhombic.

Complete step-by-step answer:So when the acid is reacted with silver nitrate the formation of silver salt occurs. The reaction of acid with silver salt is the following:
$$RCOOH + AgN{O_3} \to RCOOAg + HN{O_3}$$
So when the acid is reacted with silver nitrate the formation of silver salt and nitric acid takes place.
When the silver salt is reacted with bromine under heating the formation of bromoalkane takes place. Look at the following reaction for the understanding:
$$RCOOAg + B{r_2} \to RBr + AgBr + C{O_2}$$
So on reaction of silver salt with bromine we get bromo alkane, silver bromide and carbon dioxide.
This reaction when undergone with carbon tetrachloride and bromine with the silver salt to form bromo alkane is known as Hunsdiecker reaction. The other name for the Hunsdiecker reaction is Borodin reaction or the Hunsdiecker Borodin reaction. The use of Carbon tetrachloride is done as a solvent for the reaction to go smoothly.

Note:The Hunsdiecker reaction mechanism generally involves the organic radical intermediates. The process is firstly the reactive intermediate is formed. Then we see that to form a diradical pair the occurrence of the process decarboxylation occurs. Then the reactants are recombined to form the product.