If $M$ has a saturated elementary extension of cardinality $\lambda$, then $\lambda=\lambda^{<\lambda}$?

Question

I'm trying to prove the converse of the following statement:

$|L|<\lambda$, where $\lambda$ is such that $\lambda=\lambda^{<\lambda}$. Then every structure $M$ of cardinality $\leq \lambda$ has a saturated elementary extension of cardinality $\lambda$.

The converse would be: if $|T|<\lambda$ has a saturated model of cardinality $\lambda$, then $\lambda=\lambda^{<\lambda}$.

Here is my attempt of proof: Let $N\models T$ such that $N$ is saturated and $|N|=\lambda$. We count the possible types with parameters in subsets of $N$ of cardinality $<\lambda$. Let $A\subset N$ such that $|A|=\mu<\lambda$. Let $S_1(A)$ be the set of the complete 1-types of $N$ with parameters in $A$; then $|S_1(A)|\leq 2^\mu=\mu^\mu<\lambda^\mu$.

Moreover, $|[N]^{<\lambda}|=\lambda^{<\lambda}$, where $[N]^{<\lambda}$ is the set of the subsets of $N$ of size at most $\lambda$. Let $P=\bigcup\limits_{\mu<\lambda}\{S_1(A) \ : \ A\subset N, |A|=\mu\}$; then $|P|=\lambda^{<\lambda}\cdot \underset{A\subset N\\|A|<\lambda}{\text{sup}}|S_1(A)|=\lambda^{<\lambda}$, as $\underset{A\subset N\\|A|<\lambda}{\text{sup}}|S_1(A)|=\underset{\mu<\lambda}{\text{sup}}2^\mu<\underset{\mu<\lambda}{\text{sup}}\lambda^\mu=\lambda^{<\lambda}$.

Since $N$ must have a witness for each 1-type in $P$, then $\lambda=|N|\geq \lambda^{<\lambda}$, but in general $\lambda^{<\lambda}\geq \lambda$, and we have the equality.

Is this proof correct? Can you see other ways of proving this?

Edit: as some of you suggested, actually the converse of the statement is:

Suppose $\lambda$ is such that every structure $M$ of cardinality $\leq \lambda$ has a saturated elementary extension of cardinality $\lambda$. Then $\lambda=\lambda^{<\lambda}$.

Therefore my attempt of proof is wrong, and I close the question as @Primo Petri's answer completely satisfies the request.

Answer 1

Possibly, the correct converse of your statement is as follows:

Suppose $\lambda$ is such that every structure $M$ of cardinality $\leq \lambda$ has a saturated elementary extension of cardinality $\lambda$. Then $\lambda=\lambda^{<\lambda}$.

Let $M$ be a 2-sorted structure. One sort is $\lambda$ the other sort contains all finite maps from $\lambda$ to $\lambda$. The ternary relation $f(\alpha)=\beta$ is in the language.

Every saturated elementary extension of $M$ contains $\lambda^{<\lambda}$.