About $\lambda$-saturated ideals defined through a $\lambda$-cc forcing

In Kanamori's "The Higher Infinite" (2nd ed.) there is Lemma 17.7. When I first went through the proof it seemed obvious, but now that I'm reviewing it, I get confused.

The lemma states that if $\kappa,\lambda$ are regular cardinals, $\Bbb P$ is a $\lambda$-cc forcing and \begin{align*} \Bbb P\Vdash\dot J\text{ is a $\check V$-$\lambda$-saturated ideal over }\kappa, \end{align*} then $I=\{X\subset \kappa\mid {\Bbb P}\Vdash\check X\in\dot J\}$ is a $\lambda$-saturated ideal over $\kappa$. Here $\check V$ names the ground model, and $V$-$\lambda$-saturated means being $\lambda$-saturated if we restrict our attention to only families of sets in $V$.

The proof rests on the assertion that if we have a family $\{X_\alpha\mid \alpha<\lambda\}\subseteq \mathcal P(\kappa)$ such that for each $\alpha$ we have $X_\alpha\notin I$, then $\Bbb P\Vdash\left|\{\alpha<\lambda\mid \check X_\alpha\notin \dot J\}\right|=\lambda.$ The proof of this assertion is by contradiction, namely, if this were not the case, then there would be $\gamma<\lambda$ such that $\Bbb P\Vdash\{\alpha<\lambda\mid X_\alpha\notin \dot J\}\subseteq \check\gamma$ by $\Bbb P$ being $\lambda$-cc and $\lambda$ being regular.

I'm not understanding this step. If $\Bbb P\not\Vdash\left|\{\alpha<\lambda\mid \check X_\alpha\notin \dot J\}\right|=\lambda,$ then at best we can conclude that there is some $p\in\Bbb P$ such that $p\Vdash\left|\{\alpha<\lambda\mid \check X_\alpha\notin \dot J\}\right|<\lambda.$ However, for the contradiction to arise, this has to happen for every $p\in \Bbb P$. Why does it follow from the assumption towards contradiction that there is no $p'\in \Bbb P$ such that $p'\Vdash\left|\{\alpha<\lambda\mid \check X_\alpha\notin \dot J\}\right|=\lambda$?

We can correct Kanamori's argument as follows: we claim that there is $p\in\mathbb{P}$ such that $$p\Vdash |\{\alpha<\lambda \mid \check{X}_\alpha \notin \dot{J}\}|=\lambda.$$

Suppose not, we have $1\Vdash |\{\alpha<\lambda \mid \check{X}_\alpha \notin \dot{J}\}|<\lambda$. Since $\mathbb{P}$ is $\lambda$-c.c. and $\lambda$ is regular, $\mathbb{P}$ preserves $\lambda$ is a regular cardinal, so we can find $q$ and $\gamma<\lambda$ such that $$q\Vdash |\{\alpha<\lambda \mid \check{X}_\alpha \notin \dot{J}\}|\subseteq \check{\gamma}.$$

By using $\mathbb{P}$ being $\lambda$-c.c. and the regularity of $\lambda$, we may assume that $q=1$. (Here we use a tedious maximal antichain argument.) Now we have $1\Vdash \check{X}_\gamma\in \dot{J}$, a contradiction.

Since $\mathbb{P}$ forces $\dot{J}$ is $\check{V}$-$\lambda$-saturated, we can find $q\le p$ and $\alpha<\beta<\lambda$ such that $q\Vdash \check{X}_\alpha\cap\check{X}_\beta\notin \dot{J}$. That is, $1\not\Vdash \check{X}_\alpha\cap\check{X}_\beta\in\dot{J}$.