Interpreting the points $P_k$ as vectors, we have $P_{n+4} = \frac{P_n + P_{n+1}}{2}$, which is a linear homogeneous recursion. Its characteristic polynomial is $2x^4 - x - 1 = (x - 1)(2x^3 + 2x^2 + 2x + 1)$. So let $\alpha, \beta, \gamma$ be the roots of $f(x) = 2x^3 + 2x^2 + 2x + 1$, so that $P_n = Q_0 + Q_1 \alpha^n + Q_2 \beta^n + Q_3 \gamma^n$, $Q_0, Q_1, Q_2, Q_3$ being constant vectors.

We have $\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha) = \left(-\frac{2}{2}\right)^2 - 2 \cdot \frac{2}{2} = -1 < 0$, so $f$ has one real root and two conjugate complex roots. Suppose $\alpha$ is real. Since $f(-1) = -1 < 0$ and $f\left(-\frac{1}{2}\right) = \frac{1}{4} > 0$, $-1 < \alpha < -\frac{1}{2}$ and $|\beta| = |\gamma| = \sqrt{| -\frac{1}{2} |} < 1$, so all roots of $f$ have modulus smaller than 1 and so $P_n$ tends to $Q_0$ as $n$ goes to infinity. This means that $Q_n$ tends to $Q_0$ and thus the intersection of all $Q_n$ is $Q_0$. To find $Q_0$, notice that

$$\left| \begin{array}{l} P_0 = Q_0 + Q_1 + Q_2 + Q_3 \\ 2P_1 = 2Q_0 + Q_1 \cdot 2\alpha + Q_2 \cdot 2\beta + Q_3 \cdot 2\gamma \\ 2P_2 = 2Q_0 + Q_1 \cdot 2\alpha^2 + Q_2 \cdot 2\beta^2 + Q_3 \cdot 2\gamma^2 \\ 2P_3 = 2Q_0 + Q_1 \cdot 2\alpha^3 + Q_2 \cdot 2\beta^3 + Q_3 \cdot 2\gamma^3 \\ \Rightarrow P_0 + 2P_1 + 2P_2 + 2P_3 = 7Q_0 + f(\alpha)Q_1 + f(\beta)Q_2 + f(\gamma)Q_3 = 7Q_0 \\ \iff Q_0 = \frac{P_0 + 2P_1 + 2P_2 + 2P_3}{7} = \left(\frac{3}{7}, \frac{4}{7}\right) \end{array} \right.$$