$$\sqrt{a^2}=\sqrt{1}\ and\ \ \sqrt{b^2}\ =\ \sqrt{9}$$ $$a=\pm\sqrt{1}\ =\ \pm1\ and\ b=\pm3$$
Target question = What is the value of P?
Statement 1 $$a<0\ \ $$
So ative but b could be negative or positive
$$a=-1\ and\ \ b=\ \pm3$$ $$p=-1^4\cdot3^7=1\cdot2187=2187$$
$$OR\ p=-1^4\cdot-3^7=1\cdot-2187=-2187$$ $$So,\ P\ is\pm2187$$
Since answer is not definite
Statement 1 is NOT SUFFICIENT.
Statement 2
$$b<0,\ So\ b\ is\ negative\ $$
but a can be positive or negative $$so\ a\ \pm1$$ $$P=1^4\ \cdot-3^7=1\cdot-2187=-2187\ OR$$ $$P=-1^4\ \cdot-3^7=1\cdot-2187=-2187\ $$
So, P=-2187 Statement 2 alone is SUFFICIENT.
$$Answer\ is\ Option\ B$$
