Given A = p & 2 -3 & p+5 .

The determinant is $ad - bc$.
Given $A = \begin{pmatrix} p & 2 \\ -3 & p+5 \end{pmatrix}$.
Set the determinant to zero:

Factor the quadratic equation:

This gives two possible values for $p$:

The values of $p$ are $\boxed{-2 \text{ or } -3}$.

From the first statement, 8 shirts and 5 pairs of trousers are valued at Ksh. 49,600.

From the second statement, the number of shirts is reduced by 2 (to $8-2=6$ shirts) and the number of pairs of trousers is increased by 2 (to $5+2=7$ pairs of trousers), with a new value of Ksh. 52,800.

Let $M = \begin{pmatrix} 8 & 5 \\ 6 & 7 \end{pmatrix}$, $X = \begin{pmatrix} x \\ y \end{pmatrix}$, and $C = \begin{pmatrix} 49600 \\ 52800 \end{pmatrix}$.
We need to find $X = M^{-1}C$. First, calculate the determinant of $M$:

The price of a shirt is Ksh. 3200 and the price of a pair of trousers is Ksh. 4800.

Cost of one shirt ($x$) = Ksh. 3200. Profit on shirt = 20%.
Profit from shirt = $0.20 \times 3200 = \text{Ksh. } 640$.
Cost of one pair of trousers ($y$) = Ksh. 4800. Profit on trousers = 25%.
Profit from trousers = $0.25 \times 4800 = \text{Ksh. } 1200$.

Total cost = Cost of shirt + Cost of trousers = $3200 + 4800 = \text{Ksh. } 8000$.
Total profit = Profit from shirt + Profit from trousers = $640 + 1200 = \text{Ksh. } 1840$.

The percentage profit made from selling a shirt and a pair of trousers is $\boxed{23\%}$.