I did some computer programming to check plausibility. In future I request that you do this step yourself.
For $p_n = 3$ and $P_n = 6,$ the only prime in between is 5, and any interval of length 6 contains an integer not congruent to any prescribed value mod 5.
In C++ I was able to check up to 10,000,000. For definiteness I took the residue classes to all be 0, that is I checked multiples of the primes between $p_n$ and $P_n.$ For the $p_n$ I checked, I was able to find only relatively short intervals of consecutive numbers, each of which is divisible by at least one prime between $p_n$ and $P_n.$ That is, these intervals have lengths much shorter than $P_n$ itself. Thus in any interval of length $P_n,$ it should be quite easy to find numbers that are not divisible by any of those primes. Indeed, the probability of picking a success at random appears to increase with $p_n.$
For example, for $p_n = 5, P_n = 30,$ I tried to find long intervals where each number had at least one divisor in the set 7, 11, 13, 17, 19, 23, 29.
691558 = 2 * 7 * 47 * 1051
691559 = 11 * 62869
691560 = 2^3 * 3^2 * 5 * 17 * 113
691561 = 13 * 53197
691562 = 2 * 19 * 18199
691563 = 3 * 29 * 7949
691564 = 2^2 * 23 * 7517
691565 = 5 * 7 * 19759
The bound of 10,000,000 is not on primes, it is on the output, such as 691565 < 10,000,000.
p_n = 5 P_n = 30 0.592302 2.96151
length = 8
691558 691559 691560 691561 691562 691563 691564 691565
p_n = 7 P_n = 210 0.454539 3.18177
length = 20
635088 635089 635090 635091 635092 635093 635094 635095 635096 635097
635098 635099 635100 635101 635102 635103 635104 635105 635106 635107
p_n = 11 P_n = 2310 0.348014 3.82815
length = 43
2113 2114 2115 2116 2117 2118 2119 2120 2121 2122
2123 2124 2125 2126 2127 2128 2129 2130 2131 2132
2133 2134 2135 2136 2137 2138 2139 2140 2141 2142
2143 2144 2145 2146 2147 2148 2149 2150 2151 2152
2153 2154 2155
p_n = 13 P_n = 30030 0.283807 3.68949
length = 207
29745 29746 29747 29748 29749 29750 29751 29752 29753 29754
29755 29756 29757 29758 29759 29760 29761 29762 29763 29764
29765 29766 29767 29768 29769 29770 29771 29772 29773 29774
29775 29776 29777 29778 29779 29780 29781 29782 29783 29784
29785 29786 29787 29788 29789 29790 29791 29792 29793 29794
29795 29796 29797 29798 29799 29800 29801 29802 29803 29804
29805 29806 29807 29808 29809 29810 29811 29812 29813 29814
29815 29816 29817 29818 29819 29820 29821 29822 29823 29824
29825 29826 29827 29828 29829 29830 29831 29832 29833 29834
29835 29836 29837 29838 29839 29840 29841 29842 29843 29844
29845 29846 29847 29848 29849 29850 29851 29852 29853 29854
29855 29856 29857 29858 29859 29860 29861 29862 29863 29864
29865 29866 29867 29868 29869 29870 29871 29872 29873 29874
29875 29876 29877 29878 29879 29880 29881 29882 29883 29884
29885 29886 29887 29888 29889 29890 29891 29892 29893 29894
29895 29896 29897 29898 29899 29900 29901 29902 29903 29904
29905 29906 29907 29908 29909 29910 29911 29912 29913 29914
29915 29916 29917 29918 29919 29920 29921 29922 29923 29924
29925 29926 29927 29928 29929 29930 29931 29932 29933 29934
29935 29936 29937 29938 29939 29940 29941 29942 29943 29944
29945 29946 29947 29948 29949 29950 29951
p_n = 17 P_n = 510510 0.236611 4.0224
length = 1 + 435005 - 433756 = 1250
433756 433757 433758 433759 433760 433761 433762 433763 433764 433765
....
...
434996 434997 434998 434999 435000 435001 435002 435003 435004 435005.
Best Answer
Mauduit and Rivat showed that the sum of the binary digits of primes are equally distributed between the possibilities $0\bmod 2$ and $1\bmod 2$. So the limit you want is $1$. More generally they consider the distribution $\bmod q$ of the sums of digits of primes written in any base $b$.