The probability that the first person gets his own name is, as you say, $1/5$. The same is true for each of the others: it doesn’t matter where a person comes in the order of drawing.
There are $5!=120$ possible orders in which the $5$ slips can be drawn. I claim that exactly one-fifth of those orders have $P_2$’s slip in the second position. To see this, consider the $4!=24$ possible orders for the slips for $P_1,P_3,P_4$, and $P_5$. You can insert $P_2$’s slip into one of these orders in $5$ possible places. For instance, starting with $$P_3,P_1,P_5,P_4$$ you can get any of these:
$$\begin{align*}
&\color{red}{P_2},P_3,P_1,P_5,P_4\\
&P_3,\color{red}{P_2},P_1,P_5,P_4\\
&P_3,P_1,\color{red}{P_2},P_5,P_4\\
&P_3,P_1,P_5,\color{red}{P_2},P_4\\
&P_3,P_1,P_5,P_4,\color{red}{P_2}
\end{align*}$$
Thus, each of the $24$ possible orders for the slips for $P_1,P_3,P_4$, and $P_5$ gives you one order of all five slips in which $P_2$’s slip is first, one in which it’s second, one in which it’s third, one in which it’s fourth, and one in which it’s last. Consequently, $P_2$’s slip is equally likely to appear in any one of the five possible positions: for any position, there are $24$ orders in which it appears in that position. In particular, it appears in the second position $24$ times out of $120$, or one time in five.
Similarly, $P_3$’s slip is equally likely to appear in each position, so the probability that it’s in the third position (so that $P_3$ gets it) is $1/5$, and $P_4$ and $P_5$ also have probability $1/5$ of drawing their own slips.
Added: The probability that at least one person gets his own slip is $$1-\mathbb{P}(\text{no one gets his own slip})\;.$$ Calculating the probability that no one gets his own slip is the classic problem of derangements. It turns out that $44$ of the $120$ possible orders of drawing are derangements, i.e., orders in which no one gets his own slip, so the probability that you want is $$\frac{120-44}{120}=\frac{76}{120}=\frac{19}{30}=0.6\overline{3}\;.$$
As the number of participants increases, the probability that at least one of them gets his own slip approaches $$1-\frac1e\approx 0.63212\;.$$ As you can see, the approximation is quite good already when $n=5$.
