= Solution
Choose any positive chemical concentration scale $A_0$ and set
$$
x=L\widehat x,\qquad t=\widehat t/r,\qquad N=Kn,\qquad A=A_0a.
$$
After dropping hats, substitution gives
$$
\boxed{n_t=D_nn_{xx}-\chi(na_x)_x+n(1-n),\qquad a_t=D_aa_{xx}+\alpha n-\beta a,}
$$
where
$$
\boxed{D_n=\frac{D_N}{rL^2},\quad D_a=\frac{D_A}{rL^2},\quad\chi=\frac{\bar\chi A_0}{rL^2},\quad\alpha=\frac{\sigma K}{rA_0},\quad\beta=\frac\mu r.}
$$
For explicit dimensional checks, write bacterial-density and chemical-concentration units as $\mathsf N,\mathsf A$. Then $[D_N]=[D_A]=L^2/T$, $[r]=[\mu]=T^{-1}$, $[\bar\chi]=L^2/(T\mathsf A)$ and $[\sigma]=\mathsf A/(T\mathsf N)$. Since $[K]=\mathsf N$ and $[A_0]=\mathsf A$, each displayed ratio has unit one. The new spatial domain is $[0,1]$.
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