 Solution

Fix $i\in \{0,1\}$ and write $\kappa ={\kappa }_i$, $j=j_i$, and $M=M_i$. Every ordinal below $j(\kappa )$ is represented in the ultrapower by a function $\kappa \to \kappa$. Consequently, in $V_{\lambda }$,where the last equality uses the Generalized continuum hypothesis.

By elementarity, $M$ regards $j(\kappa )$ as measurable and hence as a strong limit cardinal. Moreover $V_{\kappa +1}\subseteq M$, so $M$ and $V_{\lambda }$ have the same subsets of $\kappa$ and the same ${\kappa }^{+}$. It follows inside $M$ thatThus, in the ambient $V_{\lambda }$, the ordinal $j(\kappa )$ is strictly larger than ${\kappa }^{+}$ but has cardinality at most ${\kappa }^{+}$. It cannot be a cardinal number. Applying this argument to both $i=0$ and $i=1$ proves that neither $j_0({\kappa }_0)$ nor $j_1({\kappa }_1)$ is a cardinal in $V_{\lambda }$, exactly as in moved critical point is not an ambient cardinal under GCH.