In Kunen's second edition of set theory he gives the following definition
Let $(\mathbb{Q},\leq_\mathbb{Q},\mathbb{1}_\mathbb{Q})$, and $(\mathbb{P},\leq_\mathbb{P},\mathbb{1}_\mathbb{P})$ be forcing posets and $i:\mathbb{Q}\rightarrow\mathbb{P}$. Then i is a complete embedding iff
a) $i(1_\mathbb{Q})=1_\mathbb{P}$.
b) $\forall{q_{1}},{q_{2}}\in{\mathbb{Q}}[q_1\leq{q_2}\rightarrow{i(q_1)\leq{i(q_2)}}]$
c) $\forall{q_{1}},{q_{2}}\in{\mathbb{Q}}[q_1\bot{q_2}\iff{i(q_1)\bot{i(q_2)}}]$
d) If $A$ is a maximal antichain of $\mathbb{Q}$, then $i(\mathbb{Q})$ is a maximal antichain of $\mathbb{P}$.
My question is why is b) not, as intuition would suggest,
b') $\forall{q_{1}},{q_{2}}\in{\mathbb{Q}}[q_1\leq{q_2}\iff{i(q_1)\leq{i(q_2)}}]$?
Is it a typo? You can't really prove b' from b as seen by $\leq_\mathbb{Q}=\{(1,q_{1}), (1,q_{2}), (1,q_{3}), (q_{1},q_{3}), (q_{2},q_{3})\}$ and $\leq_\mathbb{P}=\{(1,p_{1}), (1,p_{2}), (1,p_{3}), (p_{1},p_{2}), (p_{2},p_{3}), (p_{1},p_{3})\}$ and $i(1)=1$ and $i(q_n)=p_n$.
Indeed, (b) and (b') are not equivalent (under the assumption that only (a), (c), and (d) hold). Your counterexample is essentially correct. (Note that you used the inverse order!)
However, your main question is answered in Kunen's first edition on the pages 218 and 219 (see Definition VII.7.1 and Exercise VII.C2). Also note that Lemma III.3.72 in the second edition justifies the differences between the two editions.
Basically, (b) and (b') are equivalent under the assumption that $ \mathbb{P} $ and $ \mathbb{Q} $ are separative. (Maybe you only need that $ \mathbb{Q} $ is separative as Miha suggested in the comments.) The proof (of this Exercise VII.C2) is a routine computation.