Strongly connected and completely specified Moore equivalent of a Mealy Machine
Let $M_1 = \langle Q, \Sigma, \Delta, \delta, \lambda, q_0 \rangle $ be a Mealy machine where $\Delta$ is the output alphabet and $\lambda$ is the output function. 

Let us denote states of the corresponding Moore machine by $q_{ib}$ if there is input $a$ and output $b$ such that $\delta(q_j, a) = (q_i, b)$. This means if the Mealy machine enters a state $q$ by emitting more than two different outputs then the state $q_i$ "splits". Also, denote corresponding Moore machine's transition function by $\delta'$. 

Now consider two states $q_{ib_1}$ and $q_{jb_2}$ (of corresponding Moore machine). Existence of the state $q_{jb_2}$ implies that there is a state $q_k$ and input symbol $a$ such that $\delta(q_k, a) = (q_j, b_2)$. Since we made assumption that the Mealy machine is strongly connected, there is a string $w$ such that $\delta(q_i, w) = (q_k, b_k)$ for some output symbol $b_k$. So, we have 
$$\delta'(q_{ib_1}, wa) = \delta'(\delta'(q_{ib_1}, w),a) = \delta'(q_{kb_k}, a) = q_{jb_2}$$ 
This shows that the corresponding Moore machine is also strongly connected.

As for the completely specification, let $q_{ib_1}$ be a state of the Moore machine, and $a$ is an input symbol. Then $\delta'(q_{ib_1}, a)= (q_{jb_2}, b_2)$, where $\delta(q_i, a) = (q_j, b_2)$, which shows it is completely specified.