Solving Linear algebraic equation $P = H^* H'$ and
$H'^* H = I_2$
$P$ is $m \times m$ known matrix.
$I_2$ is $2\times 2$ identity matrix.
$H$ is $m\times 2$ matrix.
How to solve for $H$?
 A: Let 
$$
  H = \begin{bmatrix} x_1 & y_1 \\ x_2 & y_2 \\ \dots & \dots \\ x_m & y_m \end{bmatrix}
  \quad \text{and} \quad
  H' = \begin{bmatrix} x_1 & x_2 & \dots & x_m \\ y_1 & y_2 & \dots & y_m \end{bmatrix}
$$ 
Then, 
$$
  H'\cdot H = \begin{bmatrix}
             \sum_{i=1}^m x_i x_i & \sum_{i=1}^mx_i y_i \\
             \sum_{i=1}^m x_i y_i & \sum_{i=1}^m y_i y_i
             \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}
$$
If you think in terms of vectors, where
$$
  \mathbf{x} = \begin{bmatrix} x_1 & x_2 & \dots & x_m \end{bmatrix}^T
  \quad \text{and} \quad
  \mathbf{y} = \begin{bmatrix} y_1 & y_2 & \dots & y_m \end{bmatrix}^T
$$
we can write the above equation as
$$
  H'.H = \begin{bmatrix} \mathbf{x}\cdot\mathbf{x} &  \mathbf{x}\cdot\mathbf{y} \\
     \mathbf{x}\cdot\mathbf{y} & \mathbf{y}\cdot\mathbf{y} \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \,.
$$
That means the length of the vector $\mathbf{x}$ is 1, that of $\mathbf{y}$ is 1, and these two vectors are perpendicular to each other.  There can be an infinite number of such vectors.
We have to constrain the set of possible vectors ($\mathbf{x}, \mathbf{y}$) with your first equation:
$$
  P = H\cdot H' = \begin{bmatrix} 
          x_1x_1 + y_1y_1 & x_1x_2 + y_1y_2 &\dots & x_1x_m+y_1y_m \\
          x_2x_1 + y_2y_1 & x_2x_2 + y_2y_2 &\dots & x_2x_m+y_2y_m \\
          \dots & \dots & \ddots & \dots\\
          x_mx_1 + y_my_1 & x_mx_2 + y_my_2 &\dots & x_mx_m+y_my_m
        \end{bmatrix}
$$
Take the $j$-th column of matrix $P$, 
$$
  \mathbf{P}_j = x_j\,\mathbf{x} + y_j\,\mathbf{y}
$$
A vector product of this column with $\mathbf{x}$ gives
$$
  \mathbf{P}_j\cdot\mathbf{x} = x_j
$$
Expanded out,
$$
  P_{1j}\,x_1 + P_{2j}\,x_2 + \dots + P_{mj}\,x_m = x_j \,.
$$
So you have an equation that contains only $x_j$ and $P_j$ that can be written out in matrix form as
$$
  \begin{bmatrix} 
     P_{11} & P_{21} & \dots & P_{m1} \\
     P_{12} & P_{22} & \dots & P_{m2} \\
     \vdots & \vdots & \ddots & \vdots \\
     P_{1m} & P_{2m} & \dots & P_{mm}
  \end{bmatrix}
  \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_m \end{bmatrix} = 
  \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_m \end{bmatrix}  
$$ 
This is an eigenvalue problem with eigenvalue $\lambda = 1$:
$$
 (\mathbf{P}^T - \lambda\mathbf{I})\,\mathbf{x} = \mathbf{0} \,.
$$
If you repeat the process with $\mathbf{y}$, you will see that it is another eigenvector of the matrix $\mathbf{P}^T$.
I think a solution should follow at this stage.
