A good explanation is already given by arjun sir. I'm going to share my approach- (using rules of boolean algebra, checking options one by one)
$X=(P\vee Q)\rightarrow r$
$=\overline{p+q} +R$
$=\bar{P}• \bar{Q} +R$
$Y=(P\rightarrow R) \vee (Q\rightarrow R)$
$=(\bar{P} + R) + (\bar Q +R)$
$=\bar P +\bar Q + R$
A. Clearly $X\;\text {not}\equiv Y$
B. X$\rightarrow$Y $=\bar X+Y$
$=(\overline{\bar P \bar Q+R}) + (\bar P +\bar Q + R)$
$=(P+Q) \color{green}{\bar R }\; +(\bar P +\bar Q)+ \color{green}R$
$=(P+Q+R)(\bar R+R)+(\bar P+\bar Q)$
$=(P+Q+R)+(\bar P+\bar Q)$
$=1\;(TRUE)$ So option B is correct..