We'll consider only the case $C_p(A)$ ( and so $A$) nonsingular.

The $\binom{n}{p}$ minors of size $p\times p$ obtained from $p$ rows of the matrix determine the span of those $p$ rows ( Plucker coordinates). Hence we can determine the span of any $p$ rows, and by intersection, any row up to proportionality. Similarly for columns. So now we know the rows of $A$ up to proportionality, and also the columns. That is we have

$$a_{ij} = \lambda_i \cdot b_{ij} \\ a_{ij} = \mu_j \cdot c_{ij}$$

with $(b_{ij})$, $(c_{ij})$ known. To find $(a_{ij})$ (up to proportionality):

$$\frac{b_{ij}}{c_{ij}} = \frac{\mu_j}{\lambda_i}$$ that is, a matrix of rank $1$. So find $\mu_j$, $\lambda_i$ up to proportionality, and hence $(a_{ij})$.

Conclusion: $A$ is determined up to proportionality ( a $p$-th root of $1$).

Notes:

1. In the generic case, $A$ does not need to be square.

2. The case $p=n-1$ is simple since $C_{n-1} ( C_{n-1} (A)) \simeq A$ ( proportional)

3. One can recover powers of minors larger than $q$ by using the formula for the determinant of a compound matrix.

4. Using condensation formulas one can recover a fixed Hadamard power of $A$ ( up to proportionality).

We'll consider only the case $C_p(A)$ ( and so $A$) nonsingular.

The $\binom{n}{p}$ minors of size $p\times p$ obtained from $p$ rows of the matrix determine the span of those $p$ rows ( Plucker coordinates). Hence we can determine the span of any $p$ rows, and by intersection, any row up to proportionality. Similarly for columns. So now we know the rows of $A$ up to proportionality, and also the columns. That is we have

$$a_{ij} = \lambda_i \cdot b_{ij} \\ a_{ij} = \mu_j \cdot c_{ij}$$

with $(b_{ij})$, $(c_{ij})$ known. To find $(a_{ij})$ (up to proportionality):

$$\frac{b_{ij}}{c_{ij}} = \frac{\mu_j}{\lambda_i}$$ that is, a matrix of rank $1$. So find $\mu_j$, $\lambda_i$ up to proportionality, and hence $(a_{ij})$.

Conclusion: $A$ is determined up to proportionality ( a $p$-th root of $1$).

Notes:

1. In the generic case, $A$ does not need to be square.

2. The case $p=n-1$ is simple since $C_{n-1} ( C_{n-1} (A)) \simeq A$ ( proportional)

3. One can recover powers of minors larger than $q$ by using the formula for the determinant of a compound matrix.

4. Using condensation formulas one can recover a fixed Hadamard power of $A$ ( up to proportionality).

$bf{Added:}$ If $p$ and $n$ -- the size of $A$, are relatively prime, then we can get a formula for $A$ (up to proportionality. Here is the idea:

1. evey minor of size $k p$ can be determined inductively ( use Laplace expansion.

2. Get every minor of size $q p$, where $n = q \cdot p + r$, is the division of $n$ by $p$. Now, get the minors of the adjoint of $A$ of size $r$. Repeat, using Euclid's algorithm. In end we get all the minors of size $d = \operatorname{gcd}(n,p)$ of $A$ or $\operatorname{adj}(A)$. If $d=1$ we are done.