Notation: $\mathbf{A}_P = {\rm Spec}(k[P])$ with the standard log structure.

Let $n$ be an integer invertible in $k$, and consider the map $$ f_n\colon \mathbf{A}_P\to \mathbf{A}_P $$ induced by the multiplication by $n$ map $P\to P$. The map $f_n$ is log smooth (or even log etale), and hence so is its base change to the log point $\eta_P = {\rm Spec}(P\to k)$. But the fiber over $\eta_P$ is $$ {\rm Spec}(P\to k[P]/I) $$ where $I$ is the ideal generated by $n\cdot (P\setminus\{0\})$, which is non-reduced if $n>1$ and $P$ is nontrivial.

In the simplest example $P=\mathbf{N}$ you get the standard tamely ramified map $t\mapsto t^n\colon \mathbf{A}^1\to \mathbf{A}^1$.

The assumption that is often used to prevent non-reduced fibers is that the morphism is saturated. You can read about saturated morphisms in Tsuji's manuscript (now published in the Tunisian Journal of Mathematics, see Theorem II.4.2) or in Ogus' forthcoming book.

TL;DR the answer is no, but yes if(f) the morphism is saturated.

Notation: $\mathbf{A}_P = {\rm Spec}(k[P])$ with the standard log structure.

Basic counterexample. Let $n$ be an integer invertible in $k$, and consider the map $$ f_n\colon \mathbf{A}_P\to \mathbf{A}_P $$ induced by the multiplication by $n$ map $P\to P$. The map $f_n$ is log smooth (or even log etale), and hence so is its base change to the log point $\eta_P = {\rm Spec}(P\to k)$. But the fiber over $\eta_P$ is $$ {\rm Spec}(P\to k[P]/I) $$ where $I$ is the ideal generated by $n\cdot (P\setminus\{0\})$, which is non-reduced if $n>1$ and $P$ is nontrivial.

In the simplest example $P=\mathbf{N}$ you get the standard tamely ramified map $t\mapsto t^n\colon \mathbf{A}^1\to \mathbf{A}^1$.

The assumption that is often used to prevent non-reduced fibers is that the morphism is saturated. You can read about saturated morphisms in Tsuji's manuscript (now published in the Tunisian Journal of Mathematics, see Theorem II.4.2) or in Ogus' forthcoming book.