Q3.24 (d) Signals & Systems 2ed by A. Oppenheim
Let me start by the definition of the Fourier coefficients that I use:

$$f(t)=\sum_{k=-\infty}^{\infty}c_ke^{j2\pi kt/T}\tag{1}$$
with
$$c_k=\frac{1}{T}\int_{0}^Tf(t)e^{-j2\pi kt/T}dt$$

where \$f(t)\$ is periodic with period \$T\$. If you now look at the Fourier series of a Dirac comb

$$\sum_{k=-\infty}^{\infty}\delta(t-kT)=\frac{1}{T}\sum_{k=-\infty}^{\infty}e^{j2\pi kt/T}\tag{2}$$

and you compare (1) and (2), then you see that the Fourier coefficients of the signal \$y(t)\$ in your example are actually given by \$c_k=1/T\$ (for all \$k\$) with \$T=4\$, which differs from the coefficients \$b_k\$ in the example by a factor of 4. I'm not sure if this is a mistake in the book or if they use some other scaling. Anyway, I'll use the values of \$b_k\$ and \$c_k\$ as given in the example and I'll show how you can obtain \$z(t)\$. In order to get the factors right, I will use

$$y(t)=4\sum_{k=-\infty}^{\infty}\delta(t-4k)$$
which corresponds to Fourier coefficients \$b_k=1\$ for all \$k\$. From (1) the signal \$z(t)\$ is given by

$$z(t)=\sum_{k=-\infty}^{\infty}c_ke^{j2\pi kt/T}$$

Since \$c_k=0\$ for even \$k\$ we have

$$z(t)=\sum_{k\textrm{ odd}}c_ke^{j2\pi kt/T}=\sum_{k=-\infty}^{\infty}c_{2k+1}e^{j2\pi (2k+1)t/T}$$

And since \$c_{2k+1}=1\$ for all \$k\$ we can simply replace them by \$b_k\$:

$$z(t)=\sum_{k=-\infty}^{\infty}b_ke^{j2\pi (2k+1)t/T}=
\sum_{k=-\infty}^{\infty}b_ke^{j2\pi k(2t)/T}e^{2\pi t/T}=y(2t)e^{2\pi t/T}\tag{3}$$

Using \$\delta(at)=\delta(t)/|a|\$ we get

$$y(2t)=4\sum_{k=-\infty}^{\infty}\delta(2t-4k)=
4\sum_{k=-\infty}^{\infty}\delta(2(t-2k))=
2\sum_{k=-\infty}^{\infty}\delta(t-2k)\tag{4}$$

Combining (3) and (4) and using \$T=4\$ we get

$$z(t)=2\sum_{k=-\infty}^{\infty}e^{\pi t/2}\delta(t-2k)$$

which differs from the one given in your example by a factor 2. Anyway, given the definition (1) I'm pretty sure that this is correct.