### Part (a): Prove that $\triangle XYZ$ is similar to $\triangle O_1O_2O_3$

1. Identify the key points and lines:
- Let $k_1, k_2,$ and $k_3$ be circles with centers $O_1, O_2,$ and $O_3$ respectively.
- Circles $k_1$ and $k_2$ intersect at points $A$ and $P$.
- Circles $k_1$ and $k_3$ intersect at points $C$ and $P$.
- Circles $k_2$ and $k_3$ intersect at points $B$ and $P$.
- Point $X$ is on $k_1$ such that line $XA$ intersects $k_2$ at $Y$ and line $XC$ intersects $k_3$ at $Z$.

2. Establish the angles:
- Since $A$ and $P$ are points of intersection of $k_1$ and $k_2$, the line $O_1A$ is perpendicular to the tangent at $A$ and $O_2A$ is perpendicular to the tangent at $A$. Thus, $\angle O_1AP = \angle O_2AP$.
- Similarly, $\angle O_1CP = \angle O_3CP$ and $\angle O_2BP = \angle O_3BP$.

3. Use the angles to show similarity:
- Consider $\triangle O_1O_2O_3$ and $\triangle XYZ$.
- $\angle O_1O_2O_3 = \angle O_1AP + \angle O_2AP = \angle O_1AP + \angle O_1AP = 2\angle O_1AP$.
- $\angle XYZ = \angle XYA + \angle YAZ = \angle XYA + \angle XYA = 2\angle XYA$.
- Since $\angle O_1AP = \angle XYA$, we have $\angle O_1O_2O_3 = \angle XYZ$.

4. Conclude similarity:
- By the AA (Angle-Angle) criterion for similarity, $\triangle XYZ \sim \triangle O_1O_2O_3$.

$\blacksquare$

### Part (b): Prove that $P_{\triangle XYZ} \le 4P_{\triangle O_1O_2O_3}$. Is it possible to reach equality?

1. Use the similarity ratio:
- Since $\triangle XYZ \sim \triangle O_1O_2O_3$, the ratio of their areas is the square of the ratio of their corresponding sides.
- Let the ratio of the sides be $k$. Then, $P_{\triangle XYZ} = k^2 P_{\triangle O_1O_2O_3}$.

2. **Determine the maximum value of $k$:**
- The maximum value of $k$ occurs when $X, Y,$ and $Z$ are as far from $P$ as possible while still being on their respective circles.
- Since $X$ is on $k_1$, $Y$ is on $k_2$, and $Z$ is on $k_3$, the maximum distance from $P$ to these points is the diameter of the circles.
- Therefore, $k \le 2$ (since the diameter is twice the radius).

3. Calculate the maximum area ratio:
- If $k = 2$, then $P_{\triangle XYZ} = (2)^2 P_{\triangle O_1O_2O_3} = 4P_{\triangle O_1O_2O_3}$.

4. Conclude the inequality:
- Therefore, $P_{\triangle XYZ} \le 4P_{\triangle O_1O_2O_3}$.
- Equality is possible when $X, Y,$ and $Z$ are at the maximum distance from $P$ on their respective circles.

$\blacksquare$

The final answer is $P_{\triangle XYZ} \le 4P_{\triangle O_1O_2O_3}$.