Let's break down this problem step by step:

Part (a): Expand $(1 - px)^{-n}(1 + x)^6$ in ascending powers of $x$ as far as the term in $x^3$.

We will use the binomial expansion for both terms separately and then multiply them together.

Step 1: Expand $(1 - px)^{-n}$

Using the binomial expansion for $(1 - px)^{-n}$, we get:

$(1 - px)^{-n} = 1 + (-n)(-px) + \frac{-n(-n-1)}{2!}(px)^2 + \frac{-n(-n-1)(-n-2)}{3!}(px)^3 + \cdots$

Simplifying the terms:

$(1 - px)^{-n} = 1 + npx + \frac{n(n+1)}{2}p^2x^2 + \frac{n(n+1)(n+2)}{6}p^3x^3 + \cdots$

Step 2: Expand $(1 + x)^6$

Using the binomial expansion for $(1 + x)^6$, we get:

$(1 + x)^6 = 1 + 6x + \frac{6(5)}{2}x^2 + \frac{6(5)(4)}{6}x^3 + \cdots$

Simplifying the terms:

$(1 + x)^6 = 1 + 6x + 15x^2 + 20x^3 + \cdots$

Step 3: Multiply the expansions

Now, multiply the two expansions term by term, up to the $x^3$ term.

$\text{Result} = (1 + npx + \frac{n(n+1)}{2}p^2x^2 + \frac{n(n+1)(n+2)}{6}p^3x^3) \times (1 + 6x + 15x^2 + 20x^3)$

Let's expand this product term by term:

1. Constant term:

$1 \times 1 = 1$

2. Coefficient of $x$:

$1 \times 6x + npx \times 1 = 6x + npx$

So, the coefficient of $x$ is $6 + np$.

3. Coefficient of $x^2$:

$1 \times 15x^2 + npx \times 6x + \frac{n(n+1)}{2}p^2x^2 \times 1$ $= 15x^2 + 6npx^2 + \frac{n(n+1)}{2}p^2x^2$

So, the coefficient of $x^2$ is $15 + 6np + \frac{n(n+1)}{2}p^2$.

4. Coefficient of $x^3$:

$1 \times 20x^3 + npx \times 15x^2 + \frac{n(n+1)}{2}p^2x^2 \times 6x + \frac{n(n+1)(n+2)}{6}p^3x^3 \times 1$ $= 20x^3 + 15npx^3 + 6\frac{n(n+1)}{2}p^2x^3 + \frac{n(n+1)(n+2)}{6}p^3x^3$

Simplifying, the coefficient of $x^3$ is:

$20 + 15np + 3n(n+1)p^2 + \frac{n(n+1)(n+2)}{6}p^3$

Part (b): Solve for $p$ and find the coefficient of $x^3$

We are given that the coefficient of $x$ is $-17$ and the coefficient of $x^2$ is $50$.

1. For the coefficient of $x$:

$6 + np = -17$

Solving for $np$:

$np = -17 - 6 = -23$

So, $np = -23$.

2. For the coefficient of $x^2$:

$15 + 6np + \frac{n(n+1)}{2}p^2 = 50$

Substitute $np = -23$:

$15 + 6(-23) + \frac{n(n+1)}{2}p^2 = 50$ $15 - 138 + \frac{n(n+1)}{2}p^2 = 50$ $-123 + \frac{n(n+1)}{2}p^2 = 50$ $\frac{n(n+1)}{2}p^2 = 50 + 123 = 173$ $n(n+1)p^2 = 346$

Now we have two equations:

1. $np = -23$
2. $n(n+1)p^2 = 346$

From $np = -23$, solve for $p$:

$p = \frac{-23}{n}$

Substitute this into $n(n+1)p^2 = 346$:

$n(n+1)\left(\frac{-23}{n}\right)^2 = 346$ $n(n+1)\frac{529}{n^2} = 346$ $(n+1)\frac{529}{n} = 346$ $529(n+1) = 346n$ $529n + 529 = 346n$ $529 = -183n$ $n = \frac{529}{183} \approx 2.89$

Since $n$ must be a positive integer, this result suggests there may be additional constraints that need reevaluation based on rounding errors or misinterpretation of terms.

Let me correct details, but does this satisfy?