Answer
$\omega_B=90rad/s$
$\omega_A=180rad/s$
Work Step by Step
The required angular velocity can be determined as follows:
We know that
$\vec{v_D}=\vec{v_P}+\vec{\omega_B}\times \vec{r_{D/P}}~$[eq(1)]
The velocity of $D$ is given as
$\vec{v_D}=\omega_{DE}\times \vec{r_{DE}}=18\hat k\times (0.5)\hat i=9\hat j$
and $\vec{r_{D/P}}=-0.1\hat j$
We plug in the known values in eq(1) to obtain:
$9\hat j=0+(-\omega_B)\hat k\times (-0.1)\hat i=0.1\omega_B \hat j$
Comparing $j$ components on both sides, we obtain:
$9=0.1\omega_B$
$\implies \omega_B=90 rad/s$
We also know that
$\vec{v_P^{\prime}}=\vec{v_P}+\vec{\omega_B}\times \vec{r_{P^{\prime}/P}}~$[eq(2)]
As $\vec{r_{P^{\prime}/P}}=-(0.1+0.3)\hat i=-0.4\hat i$
We plug in the known values in eq(2) to obtain:
$\vec{v_{P^{\prime}}}=0+(-90\hat k)\times (-0.4\hat i)=36\hat j$
$\implies vec{v_{P^{\prime}}}=36m/s$
Now $\omega_A=\frac{v_{P^{\prime}}}{r_A}$
$\omega=\frac{36}{0.2}=180rad/s$
