Rhombus is wonderful

First notice that Points A, P, C and B all lie on the same circle. Next, notice that Points A, D, C and the centroid of $\triangle ABC$ lie on the same circle. I won't proof these since these are elementary.

Hence, the figure can be represented as: (Fig 1)

Now let's focus on triangle BPD (Fig 2)

Note that $r$ is the radius of the circles.

From Fig 2:

Through $\triangle BOP$ : $2r\cos \theta_1 = 3$

Through $\triangle POD$ and sine rule: $r\sin2\theta_1 = 2\sin \theta_2$

Also through $\triangle POD$ and cosine rule: $4+3r^2 - 8r \cos \theta_2=0$

Solving these equations give $r=\frac{\sqrt{22}}{3}$ and $\theta_1=\cos^{-1} \left(\frac{9}{2\sqrt{22}}\right)$

Referring back to Fig 1 we get

$|AP|-|PC|=2r(\sin(30° + \theta_1) - \sin(30°-\theta_1))=2r\sqrt{3}\sin\theta_1=\sqrt{\frac{7}{3}}=1.527525...$