Recovering a signal after nonuniform sampling Let $x(t)$ be a bandlimited signal such that $X(j\omega) =0 $ when $|\omega|&gt;M$. Also $p(t) = p_1(t) - p_1(t-\Delta)$ is a nonuniformly spaced periodic pulse train where $$p_1(t) = \sum_{k = -\infty}^{+\infty}\delta\left(t - \frac{2\pi k}{M}\right), \quad \text{with}\quad \Delta= \frac{\pi}{2M}$$

Let $x_p(t) = x(t)p(t)$ and apply an ideal low-pass filter with cutoff frequency $\omega_c = M$ to $X_p(j\omega)$. The result is $z(t)$. Design a system which recovers $x(t)$ from $z(t)$.

My try:
It's easy to see that
$$P(j\omega) = \left(1 - e^{-j\Delta \omega}\right)\left(M\sum_{k = -\infty}^{+\infty}\delta(\omega - kM)\right) = M\sum_{k = -\infty}^{+\infty}\left(1 - e^{-j\Delta kM}\right)\delta(\omega - kM)
$$
Also we have
$$
X_p(j\omega) = \frac{1}{2\pi}X(j\omega)\star P(j\omega) = \frac{M}{2\pi}\sum_{k = -\infty}^{+\infty}\left(1 - e^{-j\Delta kM}\right)X\big(j(\omega - kM)\big)
$$
Since $\Delta= \frac{\pi}{2M}$ we have
$$
X_p(j\omega) = \frac{M}{2\pi}\sum_{k = -\infty}^{+\infty}\left(1 - e^{\frac{-jk\pi}{2}}\right)X\big(j(\omega - kM)\big)
$$
After low-pass filter we get
$$
Z(j\omega) = \begin{cases}
\frac{M}{2\pi}\bigg[(1+j)X\big(j(\omega - M)\big) + (1-j)X\big(j(\omega + M)\big)\bigg],  &amp; \lvert \omega\rvert &lt; M\\
0, &amp; \text{O.W}
\end{cases}
$$
I've got stuck here. How can we recover $x(t)$? I tried to use phase shifter and $z(t)\cos(Mt)$ but it didn't work.
I agree with your result for $Z(j\omega)$. Apart from scaling, what you have is the original spectrum with positive and negative frequencies swapped and multiplied with factors $1-j$ and $1+j$, respectively. In order to restore the original signal, we need to add right-shifted and left-shifted versions of the spectrum, while getting rid of the complex factors. Final lowpass filtering removes the redundant components in the frequency range $[M,2M]$.
The necessary shifting and complex scaling is achieved by multiplication with
$$e^{jMt}(1+j)+e^{-jMt}(1-j)=2\cos(Mt)-2\sin(Mt)\tag{1}$$
Modulation with $(1)$, lowpass filtering with cut-off frequency $M$, and appropriate scaling restores the original signal.