How to prove the WPBE is SE in signaling game?
Claim: If choice sets $T, M,$ and $A$ are finite, then an assessment $\{\beta^*_{r}, \beta^*_{s}, \mu^*\}$ is a WPBE (weak perfect Bayesian equilibrium) of the two-stage signalling game between receiver $r$ and sender $s$ if and only if it is a SE (sequential equilibrium).

Proof: SE $\implies$ WPBE is trivial since SEs are PBEs by construction, and thus are also WPBEs.

To prove that a WPBE is also a SE for this signalling game over finite choice sets $T, M,$ and $A$, we must demonstrate that there exists a tuple $(\beta^n_r, \beta^n_s, \mu^n)$ such that $\{\beta^*_{r}, \beta^*_{s}, \mu^*\} = \lim\limits_{n \to \infty} (\beta^n_r, \beta^n_s, \mu^n)$ for some totally mixed strategy $\beta^n$ such that, for all $n$,

$$
\begin{align}
\mu^n(t|m) = \frac{\pi(t)\beta^n_s(m|t)}{\sum_{t' \in T}\pi(t')\beta^n_s(m|t')} &amp;&amp;\text{whenever} \sum_{t' \in T} \pi(t')\beta^n_{s}(m|t') &gt; 0 
\end{align}
$$ 

since this is a two-player game between players $r$ and $s$. The condition on $\mu^n$ above is known as weak (Bayesian) consistency.

To construct such a totally mixed strategy profile, $\beta^n$, consider first the following message-sending strategy profile for player $s$ as a function of $n$ conditional on the state of the world $t$:

$$
\beta^n_{s}(m|t) =
\small{
\begin{cases}
\frac{n-1}{n} \left(\frac{n-1}{n}\right) \beta^*_s(m|t) &amp;&amp; \text{if } \beta^*_s(m|t) &gt; 0 \text{ and } \sum_{t' \in T} \pi(t')\beta^*_{s}(m|t') &gt; 0 \\
\frac{n-1}{n} \left(\frac{1}{n \mathcal{N_s}(t)}\right) &amp;&amp; \text{if } \beta^*_s(m|t) = 0 \text{ and } \sum_{t' \in T} \pi(t')\beta^*_{s}(m|t') &gt; 0 \\
\frac{n-1}{n} \left(\frac{\pi(m) \mu^*(t|m)}{n\pi(t)} \right) &amp;&amp; \text{if } \mu^*(t|m) &gt; 0 \text{ and } \sum_{t' \in T} \pi(t')\beta^*_{s}(m|t') = 0 \\
\frac{1}{n} \left[1 - \sum\limits_{m' \in M} \left(\frac{n-1}{n} \frac{\pi(m') \mu^*(t|m')}{\pi(t)}\right) \right]&amp;&amp; \text{otherwise}
\end{cases}
}
$$

where, borrowing notation from Fudenberg and Tirole (1991), 

$$\mathcal{N_s}(t) \equiv \#\{m \in M\ |\ \beta^*_s(m|t) = 0 \text{ and } \sum_{t' \in T} \pi(t')\beta^*_{s}(m|t') &gt; 0 \}.$$ 

The finiteness of $M$ is invoked here so as to provide a well-defined domain over $\#\{\cdot\}$. These four cases span the parameter-space of $\beta_s^*$ and $\mu^*$ in this game. Thus $\sum_{m \in M} \beta^n_{s}(m|t) = 1$ and $\beta^n_{s}(m|t) &gt; 0$ for all $n$ and for all $m \in M$, establishing that $\beta^n_s$ is a totally mixed strategy profile for player $s$. By construction of the asymptotic behavior of $\beta^n_{s}(m|t)$, it is easy to verify that $\beta^n_{s} \to \beta^*_s$ as $n \to \infty$.

The residual strategy profile for player $r$, $\beta_r^n(a|m)$, can be simply constructed by assigning probabilities only across the parameter-space of $\beta^*_r$ since the belief $\mu^*$ is just-identified by $\beta^*_s$:

$$
\beta^n_r(a|m) = 
\begin{cases}
\left(1 - \frac{1}{n}\right)\beta^*_r(a|m) &amp;&amp; \text{if } \beta^*_r(a|m) &gt; 0 \\
\frac{1}{n\mathcal{N_r}(m)} &amp;&amp; \text{if } \beta^*_r(a|m) = 0,
\end{cases}
$$
where $\mathcal{N_r}(m) \equiv \#\{a \in A\ |\ \beta^*_r(a|m) = 0\}$. Clearly $\beta^n_r \to \beta^*_r$ as $n \to \infty$.

Lastly, by construction of $\beta^n_s(m|t)$, it is straightforward to verify that the sequence
$$
\begin{gather}
\mu^n(t|m) = \frac{\pi(t)\beta^n_s(m|t)}{\sum_{t' \in T}\pi(t')\beta^n_s(m|t')} \to \mu^*(t|m) \text{ as } n \to \infty 
\end{gather}
$$

and thus consistency is demonstrated. Hence the WPBE is also a SE for this game.$\ \blacksquare$