J.D. Lee Concise Inorganic Chemistry [3, p. 77] specifically mentions that

In case of $\ce{P(SiH3)3}$, the inter-nuclear distance is large and P atom has its own vacant d orbital. Hence the tendency to donate lone pair is very less and it adopts pyramidal structure and P atom is $\mathrm{s}p^\mathrm{3}$ hybridized.

My doubt here is:

Due to the presence of vacant d-orbital on Si atom, P atom can readily donate its lone pair and form back bond. I understand that due to large internuclear distance the chances are low.

But why would the presence of vacant d-orbital on P atom be a factor for decreasing the chances of back bonding of P atom with Si atom?

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    $\begingroup$ chemistry.stackexchange.com/questions/175368/… $\endgroup$
    – Mithoron
    Commented May 19 at 13:18
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    $\begingroup$ chemistry.stackexchange.com/questions/69568/… - that's how it works in trisilylamine - no friggin' "vacant d-orbital". It doesn't work with P for the same reason phosphines have generally smaller angle - inert pair effect. $\endgroup$
    – Mithoron
    Commented May 19 at 13:25
  • $\begingroup$ Yes, I read that but still didn't find what does the presence of vacant d-orbital on P atom has to do with reducing chances of back bonding. $\endgroup$
    – Chetan
    Commented May 19 at 13:26
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    $\begingroup$ It does not. The textbook is proposing an outdated, deprecated theory. $\endgroup$ Commented May 19 at 20:49
  • $\begingroup$ @OscarLanzi Okay, thanks but can you tell me why it is like that? $\endgroup$
    – Chetan
    Commented May 20 at 3:50