Given an isosceles right-angled triangle $ABC$ with $CA = CB = 1$, we need to find the maximum value of $PA \cdot PB \cdot PC$ where $P$ is an arbitrary point on the sides of $\triangle ABC$.

1. **Case 1: $P \in BC$**

Let $PC = x$. Then $PB = 1 - x$.

Using the Pythagorean theorem in $\triangle CPA$:
$$PA^2 = PC^2 + CA^2 = x^2 + 1 \implies PA = \sqrt{x^2 + 1}$$

Therefore, the product is:
$$PA \cdot PB \cdot PC = x(1 - x)\sqrt{x^2 + 1}$$

We need to maximize the function $h(x) = x(1 - x)\sqrt{x^2 + 1}$ for $x \in [0, 1]$.

2. **Case 2: $P \in AB$**

Let $PH \perp AC$ with $H \in AC$. Let $CH = x$. Then $HA = PH = 1 - x$.

Using the Pythagorean theorem in $\triangle CPH$:
$$PC^2 = CH^2 + PH^2 = x^2 + (1 - x)^2 = x^2 + 1 - 2x + x^2 = 2x^2 - 2x + 1$$

Since $\triangle PHA$ is a 45-45-90 triangle:
$$PA = \sqrt{2} \cdot HA = \sqrt{2}(1 - x)$$
Similarly,
$$PB = \sqrt{2} \cdot x$$

Therefore, the product is:
$$PA \cdot PB \cdot PC = 2x(1 - x) \sqrt{2x^2 - 2x + 1}$$

To find the maximum, we define:
$$f(x) = 2x(1 - x), \quad g(x) = 2x^2 - 2x + 1$$

The product function is:
$$F(x) = f(x) \sqrt{g(x)}$$

Notice that:
$$g(x) + f(x) = 2x^2 - 2x + 1 + 2x(1 - x) = 1 \implies g(x) = 1 - f(x)$$

Differentiating $F(x)$:
$$F'(x) = f'(x) \sqrt{g(x)} + f(x) \frac{g'(x)}{2\sqrt{g(x)}}$$
$$F'(x) = f'(x) \sqrt{g(x)} - f(x) \frac{f'(x)}{2\sqrt{g(x)}}$$
$$F'(x) = f'(x) \left[\sqrt{g(x)} - \frac{f(x)}{2\sqrt{g(x)}}\right]$$
$$F'(x) = f'(x) \left[\frac{2g(x) - f(x)}{2\sqrt{g(x)}}\right]$$
$$F'(x) = f'(x) \left[\frac{2(1 - f(x)) - f(x)}{2\sqrt{g(x)}}\right]$$
$$F'(x) = f'(x) \left[\frac{2 - 3f(x)}{2\sqrt{g(x)}}\right]$$

Since $\frac{2 - 3f(x)}{2\sqrt{g(x)}}$ is always positive, we have:
$$F'(x) = 0 \implies f'(x) = 0$$

Solving $f'(x) = 2 - 4x = 0$:
$$x = \frac{1}{2}$$

At $x = \frac{1}{2}$:
$$f\left(\frac{1}{2}\right) = \frac{1}{2}, \quad g\left(\frac{1}{2}\right) = \frac{1}{2}$$

Therefore:
$$F\left(\frac{1}{2}\right) = \frac{1}{2} \cdot \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{4}$$

3. **Comparison of $h(x)$ and $F(x)$**

We need to show $h(x) < F(x)$ for $x \in (0, 1)$:
$$h(x) < F(x) \implies x(1 - x)\sqrt{x^2 + 1} < 2x(1 - x) \sqrt{2x^2 - 2x + 1}$$
Dividing by the positive $x(1 - x)$:
$$\sqrt{x^2 + 1} < 2 \sqrt{2x^2 - 2x + 1}$$
Squaring both sides:
$$x^2 + 1 < 4(2x^2 - 2x + 1)$$
Simplifying:
$$x^2 + 1 < 8x^2 - 8x + 4$$
$$0 < 7x^2 - 8x + 3$$

The discriminant of $7x^2 - 8x + 3$ is:
$$D = 8^2 - 4 \cdot 7 \cdot 3 = 64 - 84 = -20 < 0$$

Therefore, $7x^2 - 8x + 3$ is always positive, confirming $h(x) < F(x)$ for $x \in (0, 1)$.

Hence, the maximum value of $PA \cdot PB \cdot PC$ for all possible positions of $P$ is $F\left(\frac{1}{2}\right) = \frac{\sqrt{2}}{4}$.

The final answer is $\boxed{\frac{\sqrt{2}}{4}}$.