There is no solution to the problem in the first version of the OP.
Proof:
We have to consider two cases:
Case a) at least one of the $P_j$ is zero
The let $k$ be the smallest index for which $P_k = 0$.
Then from
$0 = \alpha P_k = P_{k+1} + ... + P_{2k+1}$
we have
$P_{k+1} = P_{k+2} = .. = P_{2k+1} = 0$
and, inductively, $P_j = 0$ if $j \ge k$.
On the other hand
$\alpha P_{k-1} = P_{k} + ... = 0$
and so on downwards so that all $P_j = 0$ which contradicts the normalization condition. Hence we can rule out case a)
Case b) all $P_j$ are positive
I shall show that there is no solution to the recursive equations with
(1) $P_j \gt 0, j = 1, 2, 3, ...$
First from
$\alpha P_1 = P_2 + P_3$
we conclude
$\alpha \gt 0$
Notice also the $P_0$ appears only in the relation
$\alpha P_0 = P_1$
which shows that
$P_0 \gt 0$ as well, but $P_0$ does not appear in the normalization. Therefore we consider it as a mere abrevíation for $P_1/ \alpha $.
Now we transform the recursive relation into a standard form, which we define here to be one in which an element with a specific index is defined in terms of elements with smaller indices.
Define
(2) $Q_i = P_{i+1}+P_{i+2}+..., i = 0,1,2,...$
As a sum over positive quantities we have
$Q_i \gt 0, i = 0, 1, 2, ...$
The inversion of (2) is
(3) $P_i = Q_{i-1} - Q_i , i = 1, 2, ... $
Now the equations become
$\alpha P_j = Q_j - Q_{2j+1}, j =1, 2, 3, ... $
Using (3) we get
$\alpha (Q_{j-1}-Q_j) = Q_j - Q_{2j+1}$
or
(4) $Q_{2j+1} = (1+\alpha ) Q_j - \alpha Q_{j-1}, j = 1, 2, ...$
This is now a recursive relation in standard form.
The inital values are
$Q_0 = P_1 + P_2 + ... = 1$
because of the normalization condition.
And
$Q_1 = 1 - P_1 = 1 - \alpha P_0$
can be considered as a free parameter in the interval (0,1).
Before we solve (4) we observe that it defines only the elements with an odd index.
Therefore we let
$Q_{2k} = C_k > 0, k = 1, 2, ...$
with arbitrary $C_k$ in the interval (0,1).
Performing now the first few steps of the solution to (4) the reader will find that
$P_{10} = - C_5 - \alpha (1+\alpha ) Q_1$
But this is a negative quantity, and the contradiction proves the statement.