在题中等式中分别令 $n=2k-1,2k,2k+1$，$k\in\mathbb N^{\ast}$，有\[\begin{aligned}
S_{2k-1}&=-a_{2k-1}+\dfrac{1}{2^{2k-1}}+2k-4,\\
S_{2k}&=a_{2k}+\dfrac{1}{2^{2k}}+2k-3,\\
S_{2k+1}&=-a_{2k+1}+\dfrac{1}{2^{2k+1}}+2k-2,\end{aligned}\]于是\[\begin{aligned}a_{2k}&=a_{2k}+a_{2k-1}-\dfrac{1}{2^{2k}}+1,\\
a_{2k+1}&=-a_{2k+1}-a_{2k}-\dfrac{1}{2^{2k+1}}+1,\end{aligned}\]进而可得\[\begin{aligned}a_{2k-1}&=\dfrac{1}{4^k}-1,\\
a_{2k}&=3-\dfrac{1}{4^k}.\end{aligned}\]接下来考虑 $p$ 的取值范围．根据题意，$p$ 在数列 $\{a_n\}$ 的任意相邻两项之间．
一方面，有 $a_1<p<a_2$，即 $-\dfrac 34<p<\dfrac{11}4$．
另一方面，当 $-\dfrac 34<p<\dfrac{11}4$ 时，有\[a_{2k-1}-p<a_1-p<0,\]且\[a_{2k}-p>a_2-p>0,\]于是有\[\forall n\in\mathbb N^{\ast},(a_{n+1}-p)(a_n-p)<0.\]综上所述，实数 $p$ 的取值范围是 $\left(-\dfrac 34,\dfrac {11}4\right)$．