Yes, it is irrational. This is because any finite digit sequence occurs as the initial digits of a prime in your sequence (in fact you can prescribe 41% of all the digits in the beginning), hence the concatenated sequence is not periodic.

Added. The OP asked more details so here they are. Assume that $\alpha$ is rational. Then the decimal expansion of $\alpha$ is periodic after a certain digit. This implies that there is a digit $d\in\{0,\dots,9\}$ and a positive integer $n\in\mathbb{N}$ such that among any $n$ consecutive digits of $\alpha$, the digit $d$ occurs. On the other hand, by known results on the distribution of primes in arithmetic progressions, there is a $p_i$ which starts with $n+1$ digits distinct from $d$.

In fact, by a result of Huxley and Iwaniec (Mathematika 22 (1975), 188-194), the interval $(x,x+x^{0.584})$ contains a $p_i$ for any sufficiently large $x>0$. So if $x$ is sufficiently large, we can find a $p_i$ whose decimal expansion agrees with that of $x$ on the initial 41% of the digits, since the error $x^{0.584}$ only affects about the last 58.4% of the digits.

Yes, it is irrational. This is because any finite digit sequence occurs as the initial digits of a prime in your sequence (in fact you can prescribe 41% of all the digits in the beginning), hence the concatenated sequence is not periodic.

Added. The OP asked for more details so here they are. Assume that $\alpha$ is rational. Then the decimal expansion of $\alpha$ is periodic after a certain digit. This implies that there is a digit $d\in\{0,\dots,9\}$ and a positive integer $n\in\mathbb{N}$ such that among any $n$ consecutive digits of $\alpha$, the digit $d$ occurs. On the other hand, by known results on the distribution of primes in arithmetic progressions, there is a $p_i$ which starts with $n+1$ digits distinct from $d$. This is a contradiction, which shows that $\alpha$ is irrational.

In fact, by a result of Huxley and Iwaniec (Mathematika 22 (1975), 188-194), the interval $(x,x+x^{0.584})$ contains a $p_i$ for any sufficiently large $x>0$. So if $x$ is sufficiently large, we can find a $p_i$ whose decimal expansion agrees with that of $x$ on the initial 41% of the digits, since the error $x^{0.584}$ only affects about the last 58.4% of the digits.

Yes, it is irrational. This is because any finite digit sequence occurs as the initial digits of a prime in your sequence (in fact you can prescribe 47% of all the digits in the beginning), hence the concatenated sequence is not periodic.

Yes, it is irrational. This is because any finite digit sequence occurs as the initial digits of a prime in your sequence (in fact you can prescribe 41% of all the digits in the beginning), hence the concatenated sequence is not periodic.

Added. The OP asked more details so here they are. Assume that $\alpha$ is rational. Then the decimal expansion of $\alpha$ is periodic after a certain digit. This implies that there is a digit $d\in\{0,\dots,9\}$ and a positive integer $n\in\mathbb{N}$ such that among any $n$ consecutive digits of $\alpha$, the digit $d$ occurs. On the other hand, by known results on the distribution of primes in arithmetic progressions, there is a $p_i$ which starts with $n+1$ digits distinct from $d$.

In fact, by a result of Huxley and Iwaniec (Mathematika 22 (1975), 188-194), the interval $(x,x+x^{0.584})$ contains a $p_i$ for any sufficiently large $x>0$. So if $x$ is sufficiently large, we can find a $p_i$ whose decimal expansion agrees with that of $x$ on the initial 41% of the digits, since the error $x^{0.584}$ only affects about the last 58.4% of the digits.