Considering the complement of $P$ in $[1,100]$, you are asking how large can a subset of $[1,100]$ be given that it does not contain any $10$-term arithmetic progression. The more general question
How large can a set of $[1,N]$ be given that it does not contain any $k$-term arithmetic progression?
is one of the central problems in combinatorial number theory. There is no chance to give a precise answer, as an "explicit" function of $N$ and $k$, and it quite likely that this is impossible already in your special situation where $N=n^2$ and $k=n$.
Here is an argument showing that if $P\subset[1,n^2]$ meets every $n$-term progression contained in $[1,n^2]$, then $|P|>n+n^{0.5+o(1)}$. It would be interesting to improve this estimate or at least to decide whether $|P|>Cn$ holds true with an absolute constant $C>1$.
Write $K:=|P|$, $\Delta:=K-n$, and $P=\{p_1,\dotsc,p_K\}$ where $1\le p_1<\dotsb<p_K\le n^2$. Notice that $p_1\le n$ and $p_K\ge n^2-(n-1)$, whence $p_K-p_1\ge(n-1)^2$.
For any $d\in[1,n]$, the set $P$ contains an element form every residue class modulo $d$, and it follows that there are at most $K-d$ pairs of consecutive elements of $P$ with the difference equal to $d$; also, if $d>n$, then there are no such pairs at all. Let $a$ and $r$ be defined by \begin{align*} K-1 &= \Delta+(\Delta+1)+\dotsb+(\Delta+(a-1))+r \\ &= a\Delta+\frac{a(a-1)}2 + r,\quad 0\le r<\Delta+a. \tag{1} \end{align*} Since there are totally $K-1$ pairs of consecutive elements of $P$, of them at most $\Delta$ pairs at distance $n$, at most $\Delta+1$ pairs at distance $n-1$, etc, we conclude that \begin{align*} p_K-p_1 &\le n\Delta+(n-1)(\Delta+1)+\dotsb+(n-(a-1))(\Delta+(a-1))+(n-a)r \\ &= \Delta na+(n-\Delta)\cdot\frac{a(a-1)}2-\frac{a(a-1)(2a-1)}{6}+(n-a)r. \end{align*} Recalling the estimate $p_K-p_1\ge(n-1)^2$, and using ($1$), we get \begin{align*} (n-1)^2 &\le \Delta na+(n-\Delta)\cdot\frac{a(a-1)}2-\frac{a(a-1)(2a-1)}{6}+(n-a)r \\ &= n\Big(a\Delta+\frac{a(a-1)}2 + r\Big) - \Delta\cdot\frac{a(a-1)}2 - \frac{a(a-1)(2a-1)}{6} - ar \\ &= n(K-1) - \Delta\cdot\frac{a(a-1)}2 - \frac{a(a-1)(2a-1)}{6} - ar. \tag{2} \end{align*}
We now make the simplifying assumption $r=0$ and, for a contradiction, assume also that $\Delta<n^c$ with an absolute constant $0<c<0.5$. From (1) we get then $$ K-1 = \Delta a + \frac{a(a-1)}2 >\frac12\,a^2 - 1 $$ implying $a<\sqrt{2K}$; hence, $\Delta a=O(n^{0.5+c})$. As a result, $$ \frac12\,a^2 = K-1+\frac12\,a-\Delta a > K - O(n^{0.5+c}) $$ leading to $a>(1-o(1))\sqrt{2K}$.
With these estimates in mind, from (2) we obtain $$ n^2 + O(n) \le nK - \frac12\,\Delta a^2 - \frac13\,a^3; $$ that is, $$ \Delta n \ge \frac12\,\Delta a^2 + \frac13\,a^3 + O(n). $$ Consequently, $$ n^{1+c} \ge \Delta n \ge \frac13\,a^3 + O(n) \ge (1-o(1))(2K)^{1.5} + O(n) > n^{1.5} + O(n), $$ a contradiction.