Show that the Laplace smoothing for bigrams is a valid probability distribution
First, let me rewrite your formula:
$$
\Pr_L[w_j|w_i] = \frac{\#(w_i,w_j)+1}{\#w_i+|\Sigma|}.
$$
The quantity $\#w_i$ should be defined so that $\sum_j \#(w_i,w_j) = \#w_i$ (so we should ignore the final character when calculating $\#w_i$). Therefore
$$
\sum_{w_i \in \Sigma} \Pr_L[w_j|w_i] = \frac{\sum_{w_i \in \Sigma} \#(w_i,w_j) + \sum_{w_i \in \Sigma} 1}{\#w_i + |\Sigma|} = \frac{\#w_i + |\Sigma|}{\#w_i + |\Sigma|} = 1.
$$