We will show that the functions with property ($\mathcal{P}$) are precisely the constant functions on the interval $[a, b]$.

The equality in the statement can be rewritten equivalently as
$$f(x) = \frac{1}{2} \left( f\left(\frac{x+a}{2}\right) + f\left(\frac{x+b}{2}\right) \right) \quad (1)$$
We show now that, for any $n \in \mathbb{N}^*$ and any $x \in [a, b]$, the following equality holds
$$f(x) = \frac{1}{2^n} \sum_{k=0}^{2^n-1} f\left(\frac{x + (2^n - 1 - k)a + k b}{2^n}\right). \qquad (2)$$
For $n = 1$, this is precisely the relation (1). If we assume now that the equality holds for a certain positive integer $n \in \mathbb{N}^*$ and any $x \in [a, b]$, then we have
$$\begin{aligned} f(x) &= \frac{1}{2^n} \sum_{k=0}^{2^n-1} f\left(\frac{x + (2^n - 1 - k)a + k b}{2^n}\right) \\ &= \frac{1}{2^n} \sum_{k=0}^{2^n-1} \left( f\left(\frac{1}{2} \cdot \frac{x + (2^n - 1 - k)a + k b}{2^n} + \frac{1}{2} \cdot a\right) + \right. \\ & \qquad \left. + f\left(\frac{1}{2} \cdot \frac{x + (2^n - 1 - k)a + k b}{2^n} + \frac{1}{2} \cdot b\right) \right) \\ &= \frac{1}{2^{n+1}} \sum_{k=0}^{2^{n+1}-1} f\left(\frac{x + (2^{n+1} - 1 - k)a + k b}{2^{n+1}}\right). \end{aligned}$$
Consider for some positive integer $n \in \mathbb{N}^*$ and arbitrary $x \in [a, b]$ the division $\Delta_n = (x_0 = a < x_1 = \frac{(2^n-1)a+b}{2^n} < \dots < x_k = \frac{(2^n-k)a+kb}{2^n} < \dots < x_{2^n} = b)$ with norm $|\Delta_n| = \frac{b-a}{2^n}$ and the system of intermediate points
$$\xi_{(n)}(x) = \left( \xi_k(x) = \frac{x + (2^n - k)a + (k-1)b}{2^n} \mid k = \overline{1, 2^n} \right).$$
The relation (2) can then be written as
$$f(x) = \frac{1}{b-a} \cdot \sigma(f; \Delta_n, \xi_{(n)}(x)),$$
where $\sigma(f; \Delta_n, \xi_{(n)}(x))$ denotes the Riemann sum associated with the function $f$, the division $\Delta_n$ and the system of intermediate points $\xi_{(n)}(x)$. Since the function $f$ is Riemann integrable on the interval $[a, b]$, we have $\lim_{n \to \infty} \sigma(f; \Delta_n, \xi_{(n)}(x)) = \int_a^b f(s) ds$, so that $f(x) = \frac{1}{b-a} \cdot \int_a^b f(s) ds$ holds for any $x \in [a, b]$. Any function with the property $(\mathcal{P})$ is hence a constant function.

Conversely, any constant function defined on the interval $[a, b]$ is Riemann integrable and satisfies the equality in the statement.

For any $t \in \mathbb{R}$ there is then a unique function with the property ($\mathcal{P}$), namely $f : [a, b] \to \mathbb{R}$, defined by $f(x) = \frac{t}{b-a}$ for any $x \in [a, b]$, such that $\int_a^b f(x) dx = t$.