The corner points of the bounded feasible region determined by the system of linear constraints are $(0,8)$, $(4,4)$, $(12,12)$, $(0,20)$. Let $z = px + qy$, where $p, q > 0$. Condition on $p$ and $q$ so that the maximum of $z$ occurs at both the points $(12,12)$, $(0,20)$ is

The corner points of the bounded feasible region are $(0,8)$, $(4,4)$, $(12,12)$, $(0,20)$.

For the maximum of $z = px + qy$ to occur at both points $(12,12)$ and $(0,20)$, both points must give the same $z$ value, and this value must be greater than $z$ at all other points.

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At point $(12, 12)$:

$z_1 = p(12) + q(12)$

$z_1 = 12p + 12q$

At point $(0, 20)$:

$z_2 = p(0) + q(20)$

$z_2 = 20q$

For both to be maximum:

$12p + 12q = 20q$

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$12p + 12q = 20q$

$12p = 20q - 12q$

$12p = 8q$

$3p = 2q$

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With the condition $3p = 2q$, we have $p = \dfrac{2q}{3}$.

Checking $z$ values at all corner points:

At $(12, 12)$: $z = 12p + 12q = 12 \cdot \dfrac{2q}{3} + 12q = 8q + 12q = 20q$

At $(0, 20)$: $z = 0 + 20q = 20q$

At $(0, 8)$: $z = 0 + 8q = 8q$

At $(4, 4)$: $z = 4p + 4q = 4 \cdot \dfrac{2q}{3} + 4q = \dfrac{8q}{3} + \dfrac{12q}{3} = \dfrac{20q}{3}$

Since $q > 0$, we have $20q > 8q$ and $20q > \dfrac{20q}{3}$.

Therefore, both $(12,12)$ and $(0,20)$ give the maximum value.

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The condition is $3p = 2q$.