"All pentagon are polygon".
Let, Pentagon $(x):x$ is a pentagon, and Polygon $(x):x$ is a polygon
Domain of discourse: All the shapes
If $x$ is a pentagon, then $x$ is a polygon
$\forall x[\text{Pentagon}(x) \rightarrow \text{Polygon}(x)]$
Now, we can negate the above statement.
$\neg\forall x[\text{Pentagon}(x) \rightarrow \text{Polygon}(x)] \equiv \neg \forall x [\neg \text{Pentagon}(x) \vee \text{Polygon}(x)] \equiv \exists x [\text{Pentagon}(x) \wedge \neg\text{Polygon}(x)] \quad [\because \text{Using Demorgan's law:}\;\neg \forall x P(x) \equiv \exists x\neg P(x),\;\; \neg \exists x P(x) \equiv \forall x\neg P(x)]$
We can write in English language, There exist at least one pentagon, which is not a polygon. (or) there is some pentagon which is not polygon.
So, the correct answer is $(B).$