Say we have a group G, s.t P is a subgroup of G and |G|=28,|P|=7. then consider $P^g=\{gpg^{-1}|g \in G\}$ what is the order of this group ? I think that it is seven as well because if we consider the fact that the order of an element must divide the order of the group then we realise that the order of the elements must be seven or 1. it's not one as the group is too large for that so it is seven.
then if we take $(gpg^{-1})^7$ we get $(gpg^{-1})(gpg^{-1})(gpg^{-1})(gpg^{-1})(gpg^{-1})(gpg^{-1})(gpg^{-1})=gp^7g^{-1}=gg^{-1}=1$
Is this correct ?
Based on your comment, I think you meant that $P^g=\{gxg^{-1} \, | \, x \in P\}$ (also called the conjugate subgroup of $P$ via $g$.)
Let $g \in G$ be fixed and define $f:P \to G$ as $f(x)=gxg^{-1}$. Then this map is one-one, because \begin{align*} f(x)&=f(y)\\ gxg^{-1} & =gyg^{-1}\\ x&=y && (\text{left and right cancellation}) \end{align*}
Since the map is injective, therefore $|P|=|\text{Image}_f|=|P^g|$.