I couldn’t figure out a way to give small hints, so I’m afraid that I’ve actually done most of it.
Let $\kappa=\operatorname{c.c.}(\Bbb P)$, and let $\lambda=\operatorname{cf}\kappa$. For $p\in\Bbb P$ let $\alpha(p)=\sup\{|A|:A\subseteq{\downarrow\! p}\text{ and }A\text{ is an antichain}\}\;$
Suppose that $\lambda<\kappa$, and let $A=\{p\in\Bbb P:\forall q\le p(\alpha(q)=\kappa\}$. Suppose first that $A\ne\varnothing$, and let $p\in A$. Since $\alpha(p)=\kappa>\lambda$, there is an antichain $C=\{q_\xi:\xi<\lambda\}\subseteq{\downarrow\! p}$ of cardinality $\lambda$. Let $\langle\eta_\xi:\xi<\lambda\rangle$ be a sequence of cardinals cofinal in $\kappa$, and for $\xi<\lambda$ let $C_\xi$ be an antichain in ${\downarrow\! q}$ of cardinality $\eta_\xi$; use these antichains to derive a contradiction and conclude that $A=\varnothing$.
Let $D=\{p\in\Bbb P:\alpha(p)<\kappa\}$; we’ve just shown that $D$ is dense in $\Bbb P$. Let $M$ be a maximal subset of $D$ that is an antichain in $\Bbb P$; clearly $M$ is a maximal antichain in $\Bbb P$, since $D$ is dense, so $|M|<\kappa$. Suppose that $|M|<\mu<\kappa$; since $\mu<\kappa$, there is an antichain $C$ of power $\mu$, and since $\mu>|M|$, there is a $p\in M$ compatible with $\mu$ members of $C$. Let $C_p$ be the set of members of $C$ that are compatible with $p$, and for each $q\in C_p$ fix $r_q\in({\downarrow\! q})\cap({\downarrow\! p})$; clearly $\{r_q:q\in C_p\}$ is an antichain of power $\mu$ in ${\downarrow\! p}$. Thus, $\sup\{\alpha(p):p\in M\}=\kappa$, and we can recursively construct a $\lambda$-sequence $\langle p_\xi:\xi<\lambda\rangle$ in $M$ such that $\sup\{\alpha(p_\xi):\xi<\lambda\}=\kappa$ and $\alpha(p_\eta)>\sup\{\alpha(p_\xi):\xi<\eta\}$ for each $\eta<\lambda$. Now choose suitable antichains in ${\downarrow\! p_\xi}$ for $\xi<\lambda$ and piece them together to get too big an antichain in $\Bbb P$.