Let $P$ be a polynomial such that $(x-4)P(2x) = 4(x-1)P(x)$, for every real $x$. If $P(0) \neq 0$, what is the degree of $P$?

$ 
\textbf{(A)}\ 0
\qquad\textbf{(B)}\ 1
\qquad\textbf{(C)}\ 2
\qquad\textbf{(D)}\ 3
\qquad\textbf{(E)}\ \text{None of the preceding}
$
1. Given the polynomial equation:
   \[
   (x-4)P(2x) = 4(x-1)P(x)
   \]
   for every real \( x \), we need to determine the degree of \( P \) given that \( P(0) \neq 0 \).

2. First, substitute \( x = 4 \) into the equation:
   \[
   (4-4)P(8) = 4(4-1)P(4) \implies 0 = 12P(4)
   \]
   This implies \( P(4) = 0 \). Therefore, \( x = 4 \) is a root of \( P(x) \).

3. Since \( P(4) = 0 \), we can write \( P(x) \) as:
   \[
   P(x) = (x-4)Q(x)
   \]
   for some polynomial \( Q(x) \).

4. Substitute \( P(x) = (x-4)Q(x) \) into the original equation:
   \[
   (x-4)P(2x) = 4(x-1)P(x) \implies (x-4)(2x-4)Q(2x) = 4(x-1)(x-4)Q(x)
   \]
   Simplify by canceling out the common factor \( (x-4) \):
   \[
   (2x-4)Q(2x) = 4(x-1)Q(x)
   \]

5. Substitute \( x = 2 \) into the simplified equation:
   \[
   (2(2)-4)Q(4) = 4(2-1)Q(2) \implies 0 = 4Q(2)
   \]
   This implies \( Q(2) = 0 \). Therefore, \( x = 2 \) is a root of \( Q(x) \).

6. Since \( Q(2) = 0 \), we can write \( Q(x) \) as:
   \[
   Q(x) = (x-2)R(x)
   \]
   for some polynomial \( R(x) \).

7. Substitute \( Q(x) = (x-2)R(x) \) into the simplified equation:
   \[
   (2x-4)Q(2x) = 4(x-1)Q(x) \implies (2x-4)(2x-2)R(2x) = 4(x-1)(x-2)R(x)
   \]
   Simplify by canceling out the common factor \( (x-2) \):
   \[
   (2x-2)R(2x) = 4(x-1)R(x)
   \]
   Simplify further:
   \[
   2(x-1)R(2x) = 4(x-1)R(x)
   \]
   Cancel out the common factor \( 2(x-1) \):
   \[
   R(2x) = 2R(x)
   \]

8. The equation \( R(2x) = 2R(x) \) implies that \( R(x) \) must be a constant polynomial. Let \( R(x) = C \), where \( C \) is a constant.

9. Therefore, we have:
   \[
   Q(x) = (x-2)C
   \]
   and
   \[
   P(x) = (x-4)(x-2)C
   \]

10. The degree of \( P(x) \) is the sum of the degrees of its factors. Since \( (x-4) \) and \( (x-2) \) are both linear polynomials (degree 1), the degree of \( P(x) \) is:
    \[
    1 + 1 = 2
    \]

The final answer is \(\boxed{2}\).