Find the coefficient of x4 in the expansion of ((1)/(p) + px)7.

The general term in the binomial expansion of $(a+b)^n$ is given by $T_{r+1} = \binom{n}{r} a^{n-r} b^r$.
For $\left(\frac{1}{p} + px\right)^7$, we have $a = \frac{1}{p}$, $b = px$, and $n = 7$.
We want the term containing $x^4$, so we set $r=4$.

$$T_{4+1} = \binom{7}{4} \left(\frac{1}{p}\right)^{7-4} (px)^4$$ $$T_5 = \binom{7}{4} \left(\frac{1}{p}\right)^3 p^4 x^4$$

Calculate the binomial coefficient:

$$\binom{7}{4} = \frac{7!}{4!(7-4)!} = \frac{7!}{4!3!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$$

Substitute this value back into the term:

$$T_5 = 35 \cdot \frac{1}{p^3} \cdot p^4 x^4 = 35 p^{4-3} x^4 = 35p x^4$$

The coefficient of $x^4$ in the first expansion is $C_1 = 35p$.

For $\left(\frac{1}{2} - 2x\right)^6$, we have $a = \frac{1}{2}$, $b = -2x$, and $n = 6$.
We want the term containing $x^4$, so we set $r=4$.

$$T_{4+1} = \binom{6}{4} \left(\frac{1}{2}\right)^{6-4} (-2x)^4$$ $$T_5 = \binom{6}{4} \left(\frac{1}{2}\right)^2 (-2)^4 x^4$$

Calculate the binomial coefficient:

$$\binom{6}{4} = \frac{6!}{4!(6-4)!} = \frac{6!}{4!2!} = \frac{6 \times 5}{2 \times 1} = 15$$

Substitute this value back into the term:

$$T_5 = 15 \cdot \frac{1}{4} \cdot 16 x^4 = 15 \cdot 4 x^4 = 60 x^4$$

The coefficient of $x^4$ in the second expansion is $C_2 = 60$.

The problem states that the coefficient of $x^4$ in the first expansion is 192 more than the coefficient of $x^4$ in the second expansion.

$$C_1 = C_2 + 192$$

Substitute the values of $C_1$ and $C_2$:

$$35p = 60 + 192$$ $$35p = 252$$

Divide by 35 to solve for $p$:

$$p = \frac{252}{35}$$

Both 252 and 35 are divisible by 7:

$$p = \frac{252 \div 7}{35 \div 7} = \frac{36}{5}$$

The value of $p$ is $\frac{36}{5}$ or 7.2. This satisfies the condition $p > 0$.

The final answer is $\mathbf{p=\frac{36}{5}}$.