Solution:

We denote by $(X Y Z)$ and $(W X Y Z)$ the areas of $\triangle X Y Z$ and quadrilateral $W X Y Z$, respectively. We use the labels depicted in the following figure.

With equal perimeters, we get
$$a+b+z+x=c+y+z$$
or
$$a+b+x=c+y.$$
With equal areas, we get
$$(A B C)+(A C P)=(C D P)$$
Since $\triangle A C P$ and $\triangle C D P$ have the same altitude from $C$, we have
$$\frac{(A C P)}{(C D P)}=\frac{x}{y} \quad \Longrightarrow \quad (A C P)=\frac{x}{y} \cdot (C D P)$$
With $(2\star)$, we have
$$(A B C)=\left(1-\frac{x}{y}\right)(C D P)=\frac{y-x}{y} \cdot (C D P)$$
On the other hand, since $A B C D$ is cyclic, we know that $\angle D=180^{\circ}-\theta$. Then $(A B C)=\frac{1}{2} a b \sin \theta$ and $(C D P)=\frac{1}{2} c y \sin \left(180^{\circ}-\theta\right)$. After noting that $\sin \left(180^{\circ}-\theta\right)=\sin \theta$ and applying $(1\star)$, equation $(3\star)$ reduces to
$$a b=c(a+b-c)$$
This last equation is equivalent to
$$(c-b)(c-a)=0$$
which implies that $b=c$ or $a=c$.