Show that this set is a topology of X.

Let $X$ be a set with more than one point and $p\in X$. Show that $\{O:O\subset X, p \notin O \text{ or } O = X\}$is a topology of $X$.

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Your TeX didn't render on my screen and it looks a bit off. I believe you're asking about the topology that consists of sets $\tau = \{O : O \subset X, p \notin O \text{ or } O = X\}$. This is called the excluded point topology. To help you get started, I recommend taking a look at the definition of a topology. You need to show that (1) both $X$ and $\emptyset$ are in $\tau$, (2) the arbitrary union of sets in $\tau$ are in $\tau$ and (3) the intersection of two sets from $\tau$ are in $\tau$
For (1), you need to have $X \in \tau$; the TeX you have posted does not seem to include that (I see $p \notin X$ written, which seems incorrect). By my definition, $X \in \tau$. Next, is $\emptyset$ in this collection $\tau$? To answer this question, you need to ask: is $p \in \emptyset$?
For (2) consider an arbitrary union of sets from $\tau$, $\cup O_{\alpha}$. If $p \notin O_{\alpha}$ for any $O_{\alpha}$, what can we conclude about the union? You should also consider the possibility that one of the $O_{\alpha} = X$. What happens in that case?
For (3), consider $O_1, O_2$. If $p \notin O_1$ and $p \notin O_2$, what can you say about $O_1 \cap O_2$? Consider also the case that maybe $O_1 = X$. What can you say about $X \cap O_2$? Is $p$ in this intersection?