Duality in the discrete-time Fourier series
Based on the references and the definitions given, here is a proof.
Given:
$$x[n] \leftrightarrow a[k], \ and$$
$$y[n] \leftrightarrow b[k]$$
Fourier Series expansion of $x[n]e^{j\frac{2\pi}{N}nm}$ would be derived as follows:
$$\frac{1}{N}\sum^{N-1}_{n=0}x[n]e^{j\frac{2\pi}{N}nm}e^{-j\frac{2\pi}{N}nk} = \frac{1}{N}\sum^{N-1}_{n=0}x[n]e^{-j\frac{2\pi}{N}(k-m)n} = a[k-m]$$
The factor $\frac{1}{N}$ gets consumed in the definition of Fourier Series coefficients. Therefore, the following relationship can be stated:
$$x[n]e^{j\frac{2\pi}{N}nm} \leftrightarrow a[k-m]$$
Similarly, for the second case of multiplication in time-domain, you need to figure out that the inverse Fourier Series relation or the synthesis expression will be the following:
$$x[n] = \sum^{N-1}_{k=0}a[k]e^{j\frac{2\pi}{N}kn}$$
I am not adding the proof of the above which can be easily derived by plugging the expression for $a[k] = \frac{1}{N}\sum^{N-1}_{n=0}x[n]e^{-j\frac{2\pi}{N}kn}$ on the RHS.
Then all we are required to do is proving that the inverse Fourier Transform of convolution of $a[k]$ and $b[k]$ would be product $x[n]y[n]$. Which is as follows:
$$\mathcal{FS}^{-1}(a*b)[k] = \sum^{N-1}_{k=0}(a*b)[k]e^{j\frac{2\pi}{N}nk} = \sum^{N-1}_{k=0}( \sum^{N-1}_{l=0}a[l]b[k-l]) e^{j\frac{2\pi}{N}nk}$$
$$=&gt;\mathcal{FS}^{-1}(a*b)[k] =  (\sum^{N-1}_{l=0}a[l]e^{j\frac{2\pi}{N}ln})(\sum^{N-1}_{k=0}b[k-l]e^{j\frac{2\pi}{N}(k-l)n})$$
$$=&gt; \mathcal{FS}^{-1}(a*b)[k]  = \mathcal{FS}^{-1}a[k] \mathcal{FS}^{-1}b[k] = x[n]y[n]$$
I have replaced $m = (k-l)$ in the proof above to get $y[n]$.
Hence both the equations (5.69) and (5.71) are true and correctly given in the book.
I think you are trying to use equation (5.67) to prove equation (5.69) and (5.71). That is not correct, because (5.67) is stating Duality of Fourier Series expression between Continuous Time and Discrete Time. Equation (5.69) and equation (5.71) are duality of Shift-Property and Convolution Property. The second reason you cannot use Duality Expression (5.67) is because (5.67) states that if $\mathcal{FS}g[n] = f[k]$, then $\mathcal{FS}f[n] = \frac{1}{N}g[−k]$. That is it takes Fourier Series of the Fourier Series itself assuming $f[n]$ in time-domain. But in (5.69) and (5.71), the time-domain sequences remain $x[n]$ and $y[n]$ only, so (5.67) does not apply here. There are 3 Duality Properties stated in equation (5.67), (5.69) and (5.71). They are not the consequence of one another. They are a consequence of similarity in definition of Continuous time Fourier Series and Discrete Time Fourier Series.