Ans will be $-B)$
$a_{n}=P\left ( \prod _{i=1}^{\infty} X_{i}=1\right )$
$=P\left ( X_{1}=1 \right )\times P\left ( X_{2}=1 \right )\times ...........P\left ( X_{n}=1 \right )$
$=p\times p\times p........\times \infty=p^{\infty}$
Now , if $p=+ve $ number then $a_{n}\rightarrow\infty$, and if $p=fraction$, then $a_{n}\rightarrow0$, and if is $-ve$ number then $a_{n}\rightarrow-\infty$
So, it totally depends on $p$
Similarly,
$b_{n}=P\left ( \prod _{i=1}^{\infty} X_{i}=-1\right )$
$=P\left ( X_{1}=-1 \right )\times P\left ( X_{2}=-1 \right )\times ...........P\left ( X_{n}=-1 \right )$
$=p\times p\times p........\times \infty=p^{\infty}$
Same as previous
Now,
$c_{n}=P\left ( \prod _{i=1}^{\infty} X_{i}=0\right )$
$=P\left ( X_{1}=0 \right )\times P\left ( X_{2}=0 \right )\times ...........P\left ( X_{n}=0 \right )$
$=(1-2p)\times (1-2p)\times (1-2p)........\times \infty=(1-2p)^{\infty}$
Now, if $p=+ve$ integer , then value of $c_{n}$ goes to $-ve$ integer.
If $p=fraction$ , then value of $c_{n}$ goes to 0.
if $p=-ve$ integer, then value of $c_{n}$ goes to $+ve$ integer.
So, it totally depend on value of $p$