1. **State the problem:** We want to find the limit as $x \to 0$ of the expression $$\frac{1 + \sin x - \cos x}{1 + \sin(px) - \cos(px)}.$$\n\n2. **Recall the relevant formulas and approximations:** For small angles $x$, we use the Taylor series expansions:\n$$\sin x \approx x - \frac{x^3}{6}, \quad \cos x \approx 1 - \frac{x^2}{2}.$$\nSimilarly, for $px$,\n$$\sin(px) \approx px - \frac{(px)^3}{6}, \quad \cos(px) \approx 1 - \frac{(px)^2}{2}.$$\n\n3. **Apply these approximations to numerator:**\n$$1 + \sin x - \cos x \approx 1 + \left(x - \frac{x^3}{6}\right) - \left(1 - \frac{x^2}{2}\right) = x - \frac{x^3}{6} + \frac{x^2}{2}.$$\nRearranged by powers of $x$:\n$$= x + \frac{x^2}{2} - \frac{x^3}{6}.$$\n\n4. **Apply approximations to denominator:**\n$$1 + \sin(px) - \cos(px) \approx 1 + \left(px - \frac{(px)^3}{6}\right) - \left(1 - \frac{(px)^2}{2}\right) = px - \frac{p^3 x^3}{6} + \frac{p^2 x^2}{2}.$$\nRearranged:\n$$= px + \frac{p^2 x^2}{2} - \frac{p^3 x^3}{6}.$$\n\n5. **Form the ratio and simplify:**\n$$\frac{1 + \sin x - \cos x}{1 + \sin(px) - \cos(px)} \approx \frac{x + \frac{x^2}{2} - \frac{x^3}{6}}{px + \frac{p^2 x^2}{2} - \frac{p^3 x^3}{6}}.$$\nDivide numerator and denominator by $x$ (since $x \to 0$, $x \neq 0$):\n$$= \frac{1 + \frac{x}{2} - \frac{x^2}{6}}{p + \frac{p^2 x}{2} - \frac{p^3 x^2}{6}}.$$\n\n6. **Take the limit as $x \to 0$:**\nAll terms with $x$ and $x^2$ vanish, so\n$$\lim_{x \to 0} \frac{1 + \sin x - \cos x}{1 + \sin(px) - \cos(px)} = \frac{1}{p}.$$\n\n**Final answer:** $$\boxed{\frac{1}{p}}.$$