Finding the cardinality of a set I have to construct a few things before I get to my question: it's possible we don't need all of it, but I am stuck on the very last step so I should recap everything so far.
Let $\kappa$ be a singular cardinal with $\operatorname{cf}\kappa = \lambda > \omega$, and let $T \subseteq \mathcal{P}(\kappa)$. Define $T\upharpoonright_{\alpha} := \{a \cap \alpha \mid a \in T \}$. Let $C = \langle \alpha_{\xi} \mid \xi < \lambda \rangle$ be a strictly increasing continuous sequence of cardinals with limit $\kappa$.
Suppose furthermore that $\alpha < \kappa \implies \alpha^{\lambda} < \kappa$, and that the set $\{ \alpha < \kappa : | T\upharpoonright_{\alpha} | \le \alpha\}$ is stationary in $\kappa$.
Now, the set $S = \{ \xi < \lambda : | T\upharpoonright_{\alpha_{\xi}} | \le \alpha_{\xi}  \}$ is stationary in $\lambda$, essentially by hypothesis. So for each $\xi \in S$, we may define a one-to-one function $f_{\xi} : T\upharpoonright_{\alpha_{\xi} } \to \alpha_{\xi}$, and for each $a \in T$, define $g_a (\xi) = f_{\xi}(a \cap \alpha_{\xi} )$.
If we define $S_0$ to be the elements of $S$ which are limit ordinals,  then $g_a(\xi) < \alpha_{\eta}$ for some $\eta  < \xi$ by continuity of $C$, so let $h_a(\xi)$ be the least such $\eta$. Since $S_0$ is stationary,  and $h_a: S_0 \rightarrow \lambda$ is regressive, by Fodor's Theorem $h_a$ is constant on a stationary set. In other words, for each $a \in T$, there is some $S_a$ stationary $\subset S_0$ and $\eta(a) < \lambda$ such that $h_a$ is constantly $\eta(a)$ on $S_a$.
Let $(S',\eta')$ be a pair with $S' \subseteq S_0$ stationary in $\lambda$, and $\eta' < \lambda$. Define $T' = \{ a \in T \mid S_a = S' \wedge\  \eta(a) = \eta'\}$. My goal is to show that $|T'| \le \kappa$. Because $\large a \in T' \implies g_a ( \xi) < \alpha_{\eta '}$ for all $\xi \in S'$, we must have that $\large| T' \upharpoonright_{\alpha_{\xi}}|\le \alpha_{\eta '}$ for all $\xi \in S$. Now here is the part I do not understand.
"So for each $a,b \in T$ with $a \not = b$, then the sequences $\langle a \cap \alpha_{\xi} \mid \xi \in S' \rangle$ and $\langle b \cap \alpha_{\xi} \mid \xi \in S' \rangle$ are distinct, and hence $\large |T ' | \le (\alpha_{\eta '} ) ^{\lambda}$."
Now I am not sure how to verify anything in that sentence. How do I show the sequences are distinct? And how do I work out the final statement? I think that it has something to do with the cardinality of the subsets of $\large\alpha_{\eta '}$ of size $\lambda$ , because I think that set has cardinality $\large( a_{η'}  )^λ$ . Any help would be appreciated, very much  (because it means I can finish proving Silver's Theorem :).
 A: If $a,b\in T$ with $a\ne b$, there is some $\zeta\in\kappa$ that belongs to exactly one of $a$ and $b$. $C$ is cofinal in $\kappa$, so there is some $\xi\in\lambda$ such that $\zeta<\alpha_\xi$, and it follows that $a\cap\alpha_\xi\ne b\cap\alpha_\xi$. Any $\xi'\in\lambda\setminus\xi$ works equally well, $\lambda\setminus\xi$ is cofinal in $\lambda$, and $S'$ is stationary in $\lambda$, so we may as well assume that $\xi\in S'$.
In particular this is true for distinct $a,b\in T\,'$: if $a,b\in T\,'$, and $a\ne b$, there is a $\xi\in S'$ such that, and therefore the sequences $\langle a\cap\alpha_\xi:\xi\in S'\rangle$ and $\langle b\cap\alpha_\xi:\xi\in S'\rangle$ are distinct. Moreover, each of the functions $f_\xi$ is injective, so $f_\xi(a\cap\alpha_\xi)\ne f_\xi(b\cap\alpha_\xi)$ whenever $a\cap\alpha_\xi\ne b\cap\alpha_\xi$, and therefore the sequences $$\Big\langle f_\xi(a\cap\alpha_\xi):\xi\in S'\Big\rangle=\Big\langle g_a(\xi):\xi\in S'\Big\rangle$$ and $$\Big\langle f_\xi(b\cap\alpha_\xi):\xi\in S'\Big\rangle=\Big\langle g_b(\xi):\xi\in S'\Big\rangle$$ are distinct. Finally, $g_a(\xi)<\alpha_{\eta\,'}$ for all $a\in T\,'$ and $\xi\in S'$, so the map
$$T\,'\to{^{S'}\alpha_{\eta\,'}}:a\mapsto\Big\langle g_a(\xi):\xi\in S'\Big\rangle$$ is injective, and $|T\,'|\le \left|{^{S'}\alpha_{\eta\,'}}\right|=(\alpha_{\eta\,'})^{\,|S'|}=(\alpha_{\eta\,'})^\lambda$.
