No, although initially when n is low and the interval
between P(n)^2 and P(n+1)^2 causes AN to be low, NS/AN can be close to
.3. But when the interval increases, so does NS, but AN increases even
faster.

Here's some specific calculations.

n = 0 SN//AN = 0.25

n = 1 SN//AN = 0.5

n = 2 SN//AN = 0.4

n = 3 SN//AN = 0.35714285714285715

n = 4 SN//AN = 0.36363636363636365

n = 5 SN//AN = 0.23076923076923078

n = 6 SN//AN = 0.38235294117647056

n = 7 SN//AN = 0.3157894736842105

n = 8 SN//AN = 0.32608695652173914

n = 9 SN//AN = 0.29310344827586204

...

n = 100 SN//AN = 0.07769652650822668

n = 101 SN//AN = 0.07360861759425494

n = 102 SN//AN = 0.06483126110124333

n = 103 SN//AN = 0.0632688927943761

n = 104 SN//AN = 0.07355516637478109

n = 105 SN//AN = 0.07019064124783363

n = 106 SN//AN = 0.07836456558773425

n = 107 SN//AN = 0.05902192242833052

n = 108 SN//AN = 0.07178631051752922

n = 109 SN//AN = 0.064891846921797

Huge Gap from .3

...

n = 400 SN//AN = 0.02000727537286286

n = 401 SN//AN = 0.02015982564475118

n = 402 SN//AN = 0.01969642211781713

n = 403 SN//AN = 0.02106589845156644

n = 404 SN//AN = 0.019541054141269273

n = 405 SN//AN = 0.020601934790397708

n = 406 SN//AN = 0.021094029317125493

n = 407 SN//AN = 0.02070689039628704

n = 408 SN//AN = 0.01962183374955405

n = 409 SN//AN = 0.02075203973040085

n = 410 SN//AN = 0.01906106600776562

Let f[(n) be NS(n)/ AN(n) where NS(n) is number of
semiprimes in the open interval

(prime(n)^2, prime(n+1)^2) and NS(n) =
prime(n+1)^2-prime(n)^2-1 namely NS(n) is

all integers in this interval.

We have :

f(1) = 0.25

f(10) = 0.294118

f(100) = 0.218324

f(1000) = 0.169326

f(2000) = 0.162384

f(3000) = 0.157281

f(4000) = 0.153635

f(5000) = 0.151076

f(6000) = 0.149134f(7000) =
0.14768

So the answer of question is no.

I guess the limit goes to zero