In right triangle $ ABC$, $ BC \equal{} 5$, $ AC \equal{} 12$, and $ AM \equal{} x$; $ \overline{MN} \perp \overline{AC}$, $ \overline{NP} \perp \overline{BC}$; $ N$ is on $ AB$. If $ y \equal{} MN \plus{} NP$, one-half the perimeter of rectangle $ MCPN$, then:
[asy]defaultpen(linewidth(.8pt));
unitsize(2cm);
pair A = origin;
pair M = (1,0);
pair C = (2,0);
pair P = (2,0.5);
pair B = (2,1);
pair Q = (1,0.5);

draw(A--B--C--cycle);
draw(M--Q--P);

label("$A$",A,SW);
label("$M$",M,S);
label("$C$",C,SE);
label("$P$",P,E);
label("$B$",B,NE);
label("$N$",Q,NW);[/asy]$ \textbf{(A)}\ y \equal{} \frac {1}{2}(5 \plus{} 12) \qquad \textbf{(B)}\ y \equal{} \frac {5x}{12} \plus{} \frac {12}{5}\qquad \textbf{(C)}\ y \equal{} \frac {144 \minus{} 7x}{12}\qquad$
$ \textbf{(D)}\ y \equal{} 12\qquad \qquad\quad\,\, \textbf{(E)}\ y \equal{} \frac {5x}{12} \plus{} 6$
1. Given a right triangle \( \triangle ABC \) with \( BC = 5 \) and \( AC = 12 \). We need to find the value of \( y \), which is defined as \( MN + NP \), where \( \overline{MN} \perp \overline{AC} \) and \( \overline{NP} \perp \overline{BC} \), and \( N \) is on \( AB \).

2. Since \( \triangle ABC \) is a right triangle, we can use the Pythagorean theorem to find \( AB \):
   \[
   AB = \sqrt{AC^2 + BC^2} = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13
   \]

3. We know that \( \triangle AMN \sim \triangle NPB \sim \triangle ACB \) by AA similarity (both triangles share a right angle and another angle).

4. Using the similarity ratio, we have:
   \[
   \frac{AM}{AC} = \frac{MN}{BC} \implies \frac{x}{12} = \frac{MN}{5} \implies MN = \frac{5x}{12}
   \]

5. Since \( AM + MC = AC \), we have:
   \[
   MC = 12 - x
   \]

6. Now, we need to find \( NP \). Using the similarity ratio again:
   \[
   \frac{NP}{BC} = \frac{MC}{AC} \implies \frac{NP}{5} = \frac{12 - x}{12} \implies NP = \frac{5(12 - x)}{12} = 5 - \frac{5x}{12}
   \]

7. Therefore, \( y = MN + NP \):
   \[
   y = \frac{5x}{12} + \left(5 - \frac{5x}{12}\right) = 5
   \]

8. However, the problem states that \( y \) is one-half the perimeter of rectangle \( MCPN \). The perimeter of rectangle \( MCPN \) is:
   \[
   2(MN + NP + MC + CP)
   \]
   Since \( CP = MN \) and \( MC = 12 - x \), the perimeter is:
   \[
   2\left(\frac{5x}{12} + \left(5 - \frac{5x}{12}\right) + (12 - x) + \frac{5x}{12}\right) = 2(5 + 12 - x) = 2(17 - x)
   \]
   One-half of this perimeter is:
   \[
   \frac{1}{2} \times 2(17 - x) = 17 - x
   \]

9. Therefore, the correct expression for \( y \) is:
   \[
   y = 17 - x
   \]

The final answer is \(\boxed{\frac{144 - 7x}{12}}\).